INDEX
S. NO.
SUBJECT NAME
PAGE NO.
SPECIMEN COPY
.
-.
-1.
PHYSICS
1-11
2
CHEMISTRY
12 25
3
MATHEMATICS
26 40
4.
BIOLOGY
41-45
5.
SOCIAL SCIENCE
46-63
6.
MENTAL ABILITY
64-71
7.
ANSWER KEY
72-74
CLASS - X (VISTAAR-NTSE PLUS)
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PAGE # 1
FORCE
EFFECTS OF FORCE
To define force first of all one has to see the effects of force. By ‘effects of force’ we mean what force can do or what changes a force can bring about.
Effects of Force :
A force can produce the following effects : (i) A force can move a stationary body. (ii) A force can stop a moving body.
(iii) A force can change the speed of a moving body. (iv) A force can change the direction of a moving body. (v) A force can change the shape (and size) of a body.
Based on the effects of force, it may be defined as : Force is a pull or push, which changes or tends to change the state of rest or of uniform motion of a body or changes its direction or shape.
(a ) Mathematical Representation of Force :
Mathematically, force F is equal to the product of mass, m of a body and acceleration a, produced in the body due to that force.
i.e. F = ma
Where a = final velocity – initial velocity/time (b) Units of Force :
(i) In C.G.S. system :
F = ma gram × cm/s2 = dyne
If m = 1 gram, a = 1 cm/s2, then F = 1 dyne
When a force is applied on a 1 gram body and the acceleration produced in the body is 1 cm/s2 then the force acting on the body will be one dyne.
(ii) In S.I. system :
F = ma kg × m/s2 = Newton
If m = 1 kg and a = 1 m/s2 then by F = ma, F = 1 × 1 = 1 kg × m/s2 = 1 Newton.
If a force is applied on a body of mass 1 kg and acceleration produced in the body is 1 m/s2 then the force acting on the body will be one Newton.
Relationship between the newton and dyne 1 N = 1 kg × 1 m s–2 = 1000 g × 100 cm s–2 = 100000 g cm s–2 = 105 dyne Thus 1 N = 105 dyne ILLUSTRATIONS
1. A force produces an acceleration of 5.0 cm/s2 in a body of mass 20g. Then find out the force acting on the body in Newton.
Sol. Acceleration of the body, a = 5 cm/s = 0.05 m/s Mass of the body, m = 20 g = 0.02 kg
F = ma F = 0.05 × 0.02 = 10–3 N
2. A force of 15 N acts on a body of mass 5 kg for 2s. What is the change in velocity of body ?
Sol. Given : F = 15 N , t = 2s , m = 5 kg F = ma a = m F = 5 15 = 3 m/s2 a = t u v v – u = at = 3 × 2 = 6m/s RESULTANT FORCE
Many forces may be simultaneously applied on a body, for example- several persons may jointly make an effort to move a heavy body, each person pushes it i.e. each person applies a force on it. t is also possible that a stronger man pushes that body hard enough and produces same acceleration in it. f a single force acting on a body produces the same acceleration as produced by a number of forces, then that single force is called the resultant force of these individual forces .
BALANCED AND UNBALANCED FORCES
Suppose a toy car can move on its wheels along east-west direction. If we push the car with our left hand, it moves towards east. If we push the car with our right hand, it will move towards west. If we push the car with our left hand towards east and with right hand towards west simultaneously, it is possible that the car will remain at rest.n this case the forces balance each other and there is no acceleration.
So if a set of forces acting on a body produces no acceleration in it, the forces are said to be balanced forces.f it produces a non-zero acceleration, the forces are said to be unbalanced. If two forces balance each other, they must be in opposite direction and have equal magnitudes.
PAGE # 2 (a) Definit ion of Ba lanc ed Force s :
When two forces of equal magnitude are acting in opposite directions on an object simultaneously, then the object continues in its state of rest or of uniform motion in a straight line. Such forces acting on the object are known as balanced forces.
(b) De finition of Unba la nc ed Forc es :
When two forces of unequal magnitudes are acting in opposite directions on an object simultaneously, then the object moves in the direction of a large force. These forces acting on the object are known as unbalanced forces.
SOME COMMON FORCES
(a) Contact Force :
When a body A is in contact with body B, A can exert force on B and B can exert force on A, these forces are called contact forces. Push or pull by a person, force by wind, force by a weight on the head of a porter etc, are the examples of contact forces.
(b) Normal Forc e:
If the contact forces between the bodies are perpendicular to the surfaces in contact, then the forces are known as normal forces.
Eg. : Consider a book on a table. The table pushes the book upwards and book pushes the table downwards, these forces are perpendicular to the surfaces of book and table. Thus the table applies a normal force on book in the upward direction and book applies a normal force on table in downward direction.
(c) Force of Friction :
Two bodies placed in contact can also exert forces parallel to the surfaces in contact, such a force is called force of friction or simply friction.
force by earth
table
force of friction
normal force by table
applied force
Suppose a body is placed on the table following three forces act on it :
Force by earth in downward direction.
Normal force due to table in upward direction. Applied force towards right.
Body is not moving, so all the forces must be balanced. Normal force by table and force by earth are balanced with each other.To balance the applied force there must be an equal and opposite force. This force in known
as force of friction. If we increase the applied force and the body is still at rest, it means force of friction also increases till it is balanced by the applied force. The force of friction is self adjusting force. On increasing the applied force the force of friction will increase upto a limit. t is known as limiting friction after it on increasing the applied force, the body starts to move.
Force of friction is of two types. (i) Static Friction :
The magnitude of static friction is equal and opposite to the external force exerted, till the object at which force is exerted, is at rest. This means it is a variable and self adjusting force. However it has a maximum value called limiting friction.
fs(max) = s N
The actual force of static friction may be smaller than sN and its value depends on other forces acting on
the body. the magnitude of frictional force is equal to that required to keep the body at relative rest.
0
fs
fs(max) (ii) Kinetic Friction :The magnitude of the kinetic friction is proportional to the normal force acting between the two bodies. We can write
fk N fk = k N
Where N is normal force. The proportionality constant k is called the coefficient of kinetic friction and its value
depends of the nature of the two surfaces in contact. If the surfaces are smooth k will be small, if the surfaces are rough k will be large. It also depends on the materials of the two bodies in contact.
NOTE :
Here s and k are proportionality constants. s is called
coefficient of static friction and k is called coefficient
of kinetic friction. They are dimensionless quantities and independent of shape and area of contact. It is a property of the two contact surfaces.
s is always greater then k for a given pair of surfaces.
If it is not mentioned, then s = k can be taken. The value of can be from 0 to1.
Following graph shows the variation of frictional force with applied force.The actual value of depends on the degree of smoothness and other environmental factors.
Eg. : Wood may be prepared at various degress of smoothness and the coefficient of friction will vary.
PAGE # 3 (d) Force Exert ed by Spring :
A spring is made of a coiled metallic wire having a definite length. When it is neither pushed nor pulled then its length is called natural length. At natural length the spring does not exert any force on the objects attached to its ends.f the spring is pulled at the ends, its length becomes larger than its natural length, it is known as stretched or extended spring. Extended spring pulls objects attached to its ends.
A B Normal spring A B Force on A Force on B Stretched spring
A
B
Force on A
Force on B
Compressed spring
If the spring is pushed at the ends, its length becomes less than natural length. It is known as compressed spring. A compressed spring pushes the objects attached to its ends.
F x F F x F F = 0 spring in natural length does not exerts any force on its ends
F = – kx x = compression in spring F = – kx ;k = spring constant or stiffness constant (unit = N/m) x = extension in spring Fext Fext (e) Weight :
The earth attracts all the bodies towards its centre.The force exerted by the earth on the body is known as the weight of the body .It acts in vertically downward direction. These forces are not contact forces. If mass of the body is m and gravitational acceleration is g, then the weight of the body will be mg, here g = 9.8 m/s2.
(f) Tension :
Tension is the magnitude of pulling force exerted by a string, cable, chain, rope etc. W hen a string is connected to a body and pulled out, the string said to be under tension. It pulls the body with a force T, whose direction is away from the body and along the length of the string. Usually strings are regarded to be massless and unstretchable, known as ideal string.
Note : (i) Tension in a string is an electromagnetic
force and it arises only when string is pulled. If a massless string is not pulled, tension in it is zero. (ii) String can not push a body in direct contact.
CONSERVATIVE AND NON-CONSERVATIVE FORCE
( a) C ons erv at ive Forc e :
A force is said to be conservative if the amount of work done in moving an object against that force is independent of how the object moves from the initial position to the final position.
One important example of conservative force is the gravitational force. It means that amount of work done in moving a body against gravity from location A to location B is the same whichever path we may follow in going from A to B. This is illustrated in figure.
m
m
B
A
h
A force is conservative if the total work done by the force on an object in one complete round is zero, i.e. when the object moves around any closed path (returning to its initial position).
This definition illuminates an important aspect of a conservative force viz. Work done by a conservative force is recoverable. Thus in figure, we shall have to do mgh amount of work in taking the body from A to B. However, when body is released from B, we recover mgh of work.
Other examples of conservative forces are spring force, electrostatic force etc.
( b) Non-C ons e rv a t iv e Forc e :
A force is non-conservative if the work done by that force on a particle moving between two points depends on the path taken between the points.
PAGE # 4 The force of friction is an example of non-conservative
force. Let us illustrate this with an instructive example. Suppose we were to displace a book between two points on a rough horizontal surface (such as a table). If the book is displaced in a straight line between the two points, the work done by friction is simply FS where :
F = force of friction ;
S = distance between the points.
However, if the book is moved along any other path between the two points (such as a semicircular path), the work done by friction would be greater than FS. Finally, if the book is moved through any closed path, the work done by friction is never zero, it is always negative. Thus the work done by a non-conservative force is not recoverable, as it is for a conservative force.
GALILEO’S EXPERIMENTS
Experiment 1 :
It was observed by Galileo that when a ball is rolled down on an inclined frictionless plane its speed increases, whereas if it is rolled up an inclined frictionless plane its speed decreases .If it is rolled on a horizontal frictionless plane the result must be between the cases describe above i.e. the speed should remain constant. It can be explain as :
v’
v
v’ = v
moving down : speed increases moving up : speed decreases moving horizontal : speed remains constant Experiments 2 :
When a ball is released on the inner surface of a smooth hemisphere, it will move to the other side and reach the same height before coming to rest momentarily. f the hemisphere is replaced by a surface shown in figure(b) in order to reach the same height the ball will have to move a larger distance.
(a) h (b) h (c) v v
If the other side is made horizontal, the ball will never stop because it will never be able to reach the same height, it means its speed will not decrease. It will have uniform velocity on the horizontal surface. Thus, if unbalanced forces do not act on a body, the body will either remain at rest or will move with a uniform velocity. It will remain unaccelerated.
Newton concluded the idea suggested by Galileo and was formulated in the laws by Newton.
NEWTON’S FIRST LAW OF MOTION
Every body remain in its state of rest or uniform motion in a straight line unless it is compelled by some external force.
It means a body remain unaccelerated if and only if, the resultant force on it is zero.
In such a case the body is said to be in equilibrium.
INERTIA
(a ) De finition of Ine rtia :
The tendency of the body to oppose the change its states of rest or uniform motion in a straight line is called inertia. Newton’s first law of motion is also called law of inertia.
( b) D e s c r ipt ion :
It follows from first law of motion that in absence of any external force, a body continues to be in its state of rest or in uniform motion along a straight line. In other words, the body cannot change by itself its position of rest or of uniform motion.
(c ) Ine rt ia De pe nds upon Mas s :
We know that it is difficult to move a heavier body than the lighter one. Similarly it is difficult to stop a moving heavier body than a lighter body moving with the same velocity. Thus, we conclude that mass of the body is the measure of inertia, more the mass, more the inertia.
TYPES OF INERTIA
There are three types of Inertia which are : (a ) Inert ia of Rest :
The tendency of the body to oppose the change in its state of rest when some external unbalance force is applied on it, is called the inertia of rest.
PAGE # 5 Example based on Inertia of rest :
A person sitting in a bus falls backwards when the bus suddenly starts. The reason is that lower part of his body begins to move along with the bus but the upper part of his body tends to remain at rest due to inertia of rest.
(b) Inert ia of Mot ion :
The tendency of the body to oppose its state of motion when some unbalance forces are applied on it, is called the inertia of motion.
Example based on Inertia of motion :
A man carelessly getting down a moving bus falls forward, the reason being that his feet come to rest suddenly, whereas the upper part of his body retains the forward motion.
(c) Inertia of Direction :
The tendency of a body to oppose any change in its direction of motion is known as inertia of direction. Example based on Inertia of direction :
Tie a stone to one end of a string and holding other end of the string in hand, rotate the stone in a horizontal circle. If during rotation, the string breaks at certain stage, the stone is found to fly off tangentially at that point of the circle.
String Breaks
String breaks, stone goes away tangentially
Definition of force from first law of motion :
According to first law of motion, if there is no force, there is no change in state of rest or of uniform motion. In other words, if a force is applied, it may change the state of rest or of uniform motion. If the force is not sufficient, it may not produce a change but only try to do so. Hence force is that which changes or tries to change the state of rest or of uniform motion of a body in straight line.
MOMENTUM D e finit ion :
Momentum of a particle may be defined as the quantity of motion possessed by it and it is measured by the product of mass of the particle and its velocity. Momentum is a vector quantity and it is represented by p mv p Unit of mome nt um :
(In C.G.S. system) p = mv gram × cm/s = dyne × s (In M.K.S. system)p = mv kg × m/s = Newton × s
3. A ball of mass 100 gm. is moving with a velocity of 15 m/s. Calculate the momentum associated with the ball.
Sol .
Mass of the ball = 100 gm. = 1000
100 kg. = 0.1 kg.
Velocity of the ball = 15 m/s
So, momentum = mass of the ball × velocity of the ball
= 0.1 kg. × 15 m/s = 1.5 kg. m/s
PAGE # 6
NEWTON’S SECOND LAW OF MOTION
The rate of change of momentum of a body is directly proportional to the applied unbalanced forces i.e. Rate of change of momentum Force applied
Let a body is moving with initial velocity u and after applying a force F on it, its velocity becomes v in time t.
Initial momentum of the body p1 = mu Final momentum of the body p2 = mv Change in momentum in time t is mv – mu
So rate of change of momentum = t
mu – mv
But according to Newton’s second law, t mu – mv F or F t ) u – v ( m Here, t u – v = a (acceleration) So Fma
or F = kma (Here k is proportionality constant. If 1N force is applied on a body of mass 1 kg and the acceleration produced in the body is 1 m/s2, then 1 = k × 1 × 1 or k = 1
Hence, F = ma
So the magnitude of the resultant force acting on a body is equal to the product of mass of the body and the acceleration produced. Direction of the force is same as that of the acceleration.
UNITS OF FORCE
( a ) In C.G.S. System :
F = ma gm × cm/s2 = Dyne Definition of one dyne :
If m = 1 gm, a = 1 cm/s2, then F = 1 dyne.
When a force is applied on a body of mass 1 gram and the acceleration produced in the body is 1 cm/s2, then the force acting on the body will be one dyne.
(b) In S.I. System : F = ma kg × m/s2 = Newton Definition of one Newton :
If m = 1 kg and a = 1 m/s2 then by, F = ma F = 1 × 1 = 1 kg × m/s2 = 1 N.
If a force is applied on a body of mass 1 kg and acceleration produced in the body is 1 m/s2, then the force acting on the body will be one Newton.
(c) Kilogram Force (kgf) :
Kilogram force (kgf) or Kilogram weight (kg. wt.) is force with which a mass of 1 kg is attracted by the
earth towards its centre.
1kgwt = 1kgf = 9.8 N
(d) Gram Forc e (gf) :
Gram force or gram weight is the force with which a mass of 1 gram is attracted by the earth towards its centre.
1gwt = 1gf = 981 dyne
Abou both the units are called gravitational unit of force. Relation between Newton and dyne.
We know : 1 N = 1kg × 1ms-2
or 1 N = 1000 g × 100 cms-2 or 1 N = 105 g cms-2 = 105 dyne 1 N = 105 dyne
4. A force of 20N acting on a mass m1 produces an acceleration of 4 ms–2. The same force is applied on mass m2 then the acceleration produced is 0.5 ms–2. What acceleration would the same force produce, when both masses are tied together ?
Sol. For mass m1: F = 20N, a = 4 ms–2 then m1 = a F = 4 20 = 5 kg For mass m2 : F = 20N, a = 0.5 ms–2 then m2 = a F = 5 . 0 20 = 40 kg
When m1 and m2 are tied together : Total mass = m1 + m2 = 45 kg, F = 20N then a = ) m m ( F 2 1 = 45 20 = 0.44 ms–2 IMPULSE OF FORCE
A large force acting for a short time to produce a finite change in momentum is called impulsive force. The product of force and time is called impulse of force. i.e., Impulse = Force × Time
or I = Ft
The S.I. unit of impulse is Newton-second (N-s) and the C.G.S unit is dyne- second (dyne-s)
Im pu ls e a nd M ome nt um :
From Newton’s second law of motion
Force, F = t p p2 1 or Ft = p2– p1 i.e., Impulse = Change in momentum
This relation is called impulse equation or momentum-impulse theorem. It has an important application in our everyday life.
PAGE # 7
IMPULSE DURING AN IMPACT OR COLLISION The impulsive force acting on the body produces a change in momentum of the body on which it acts. We know, Ft = mv – mu, therefore the maximum force needed to produce a given impulse depends upon time. If time is short, the force required in a given impulse or the change in momentum is large and vice-versa.
NEWTON’S THIRD LAW (a) Statement :
The law states that “ To every action there is an equal and opposite reaction“. Moreover, action and reaction act on different bodies.
( b) D emonst rat ion :
Two similar spring balances A and B joined by hook as shown in the figure. The other end of the spring balance B is attached to a hook rigidly fixed in a rigid wall.
Demonstration- Newton’s third law of motion
The other end of the spring balance A is pulled out to the left. Both balances show the same reading (20 N) for the force.
The pulled balance A exerts a force of 20N on the balance B. It acts as action, B pulls the balance A in opposite direction with a force of 20 N. This force is known as reaction.
We conclude that action-reaction forces are equal and opposite and act on two different bodies.
NO ACTION IS POSSIBLE WITHOUT REACTION
Examples :
(i) A nail cannot be fixed on a suspended wooden ball. (ii) A paper cannot be cut by scissors of single blade. (iii) A hanging piece of paper cannot be cut by blade. (iv) Writing on a hanging page is impossible. (v) Hitting on a piece of sponge does not produce reaction. You do not enjoy hitting.
ACTION AND REACTION ARE NOT BALANCED Action and reaction, though equal and opposite are not balanced because they act on two different bodies. In case when they act on two different bodies forming a single system, they become balanced.
ANY PAIR OF EQUAL AND OPPOSITE FORCES IS NOT AN ACTION–REACTION PAIR
Consider a book kept on a table. We have seen that the table pushes the book in the upward direction. Then why does not the book fly up? It does not fly up because there is another force on the book pulling it down. This is the force exerted by the earth on the book, which we call the weight of the book. So, there are two forces on the book– the normal force, N acting upwards, applied by the table and the force, W acting downwards, applied by the earth. As the book does not accelerate, we conclude that these two forces are balanced. In other words, they have equal magnitudes but opposite directions. V V V
N
N=W
Can we call N the action and W the reaction ? We cannot. This is because, although they are equal and opposite, they are not forces applied by two bodies on each other. The force N is applied by the table on the book, its reaction will be the force applied by the book on the table. Weight W is the force applied by the earth on the book, its reaction will be the force applied by the book on the earth.
So, although N and W are equal and opposite, they do not form an action–reaction pair.
PRINCIPLE OF CONSERVATION OF LINEAR MOMENTUM
By Newton’s second law, the rate of change of momentum is equal to the applied force.
time momentum in Change = Force Change in momentum = F × t If F = 0 then, Change in momentum = 0
If the force applied on the body is zero then its momentum will be conserved, this law is also applicable on the system. If in a system the momentum of the objects present in the system are P1, P2, P3... and external force on the system is zero, then– P1 + P2 + P3 +... = Constant
NOTE : If only internal forces are acting on the system then its linear momentum will be conserved.
PAGE # 8 ( a ) The Law of Conservation of Linear
Momentum by Third Law of Motion :
Suppose A and B are two objects of masses m1 and m2 are moving in the same direction with velocity u1 and u2 respectively (u1 > u2). Object A collides with object B and after time t both move in their original direction with velocity v1 and v2 respectively.
The change in momentum of object A = m1v1– m1u1
u1 u2 m1 m2 before collision (u > u )1 2 The force on B by A is F1 = time momentum in Change F1 = t u m – v m1 1 1 1 ...(1)
The change in momentum of object B = m2v2– m2u2
The force on A by B is F2 = time momentum in Change = t u m – v m2 2 2 2 ...(2) v1 v2 m1 m2 after collision By Newtons third law, F1 = –F2
t u m – v m1 1 1 1 = t u m – v m – 2 2 2 2
m1v1– m1u1 = –m2v2 + m2u2 or m1u1 + m2u2 = m1v1 + m2v2or Initial momentum = Final momentum
SOME ILLUSTRATION ON CONSERVATION OF MOMENTUM
(a) Recoil of Gun :
A loaded gun (rifle) having bullet inside it forming one system is initially at rest. The system has zero initial momentum.
v V
When the trigger (T) is pressed, the bullet is fired due to internal force of explosion of powder in cartidge inside. The bullet moves forward with a high velocity and the gun move behind (recoils) with a lesser velocity.
Let the bullet and the gun have masses m and M respectively. Let the bullet move forward with velocity v and the gun recoils with velocity V.
Then final momentum of the gun and bullet is MV + mv By the law of conservation of momentum–
Initial momentum of the system = Final momentum of the system.
0 = MV + mv
or V = – M mv
Hence the recoil velocity of gun = M mv
and the velocity of the gun is = – M mv
(b) The Working of a Rocke t :
the momentum of a rocket before it is fired is zero. When the rocket is fired, gases are produced. These gases come out of the rear of the rocket with high speed. The direction of the momentum of the gases coming out of the rocket is in the downward direction. Thus, to conserve the momentum of the system i.e., (rocket + gases), the rocket moves upward with a momentum equal to the momentum of the gases. So, the rocket continues to move upward as long as the gases are ejected out of the rocket. Thus a rocket works on the basis of the law of conservation of momentum.
TRANSLATORY EQUILIBRIUM
When several forces acts on a body simultaneously in such a way that resultant force on the body is zero, i.e.,
F= 0 with F=
i
Fthe body is said to be in translatory equilibrium. Here it is worthy to note that :
(i) As if a vector is zero all its components must vanish i.e. in equilibrium as -F= 0 with F =
i F= 0 xF
= 0 ;
Fy= 0 ; zF
= 0So in equilibrium forces along x axis must balance each other and the same is true for other directions. If a body is in translatory equilibrium it will be either at rest or in uniform motion. If it is at rest, equilibrium is called static, otherwise dynamic.
Static equilibrium can be divided into following three types :
( a ) St a ble e quilibrium :
If on slight displacement from equilibrium position a body has a tendency to regain its original position it is said to be in stable equilibrium. In case of stable equilibrium potential energy is minimum and so center of gravity is lowest.
PAGE # 9 O
(b) Unstable equilibrium : If on slight displacement from equilibrium position a body moves in the direction of displacement, the equilibrium is said to be unstable. In this situation potential energy of body is maximum and so center of gravity is highest.
O
(c) Neutral equilibrium : If on slight displacement from equilibrium position a body has no tendency to come back to its original position or to move in the direction of displacement, it is said to be in neutral equilibrium. In this situation potential energy of body is constant and so center of gravity remains at constant height.
CENTRE OF MASS (C.M.)
It is the point inside or outside the body at which the whole mass of the body is supposed to be concentrated. If an external force applied on the centre of mass, it will produce the same motion in the body as if the body is a point mass.
CENTRE OF GRAVITY (C.G.)
It is a point inside or outside the body at which the whole weight of the body is supposed to be acting. If an external force is applied on the centre of gravity, it will make the body move in the direction of the force just as a particle moves.
IMPORTANT NOTE :
For bodies of regular shape having uniform density the C.M and the C.G. lie at the geometrical centre of the body.
Example :
For Ring : The centre of the ring (it lies out side the body)
For Rectangle or Square : At the point of intersection of its diagonals.
For cylinder : At the centre of the axis.
TORQUE
If a body is free to move about an axis and a force is applied on the body then it rotates about that axis. The capability of the force to rotate the body or to change the rotational motion of the body is known as torque.
Torque of force F about the axis passing through the point O is
Torque = force × perpendicular distance
M
P
O
r
F
= F(OM) From OMP sin = r OM , OM = rsinHence, = Fr sin ...(i)
Case-I If = 0º
From equation (i)= Fr sin 0º= 0
Case-II If = 90º
From equation (i)= Fr sin 90ºFr max. Case-III If = 180º
From equation (i)= Fr sin 180º = 0
EXERCISE
1. A force of 10 N is applied on a body of 2 kg mass at rest . The distance travelled by the body in 2 sec. is: (A) 10 m (B) 20 m
(C) 30 m (D) 40 m
2. A body of 10 N weight is resting on a plane surface . If external force of 5 N is exerted on the body as shown in the figure , then friction force acting on the body by surface is :
(A) 0 (B) 5 N
(C) 10 N (D) 15 N 3. When a body is stationary :
(A)there may be no force acting on it (B)there may be one force acting on it (C)there may be couple of forces acting on it (D)the body is in vacuum
4. A body of mass 'm' and velocity ‘u’ strikes a wall and rebounds with a velocity ‘v’. The change in momentum is
(A) m (v – u) (B) m (u – v)
PAGE # 10
5. A constant force acts on a body of mass m at rest for t seconds and then ceases to act. In next t sec-onds the body travels a distance x, magnitude of force is. (A) 2 t mx (B) t mx (C) mxt (D) mxt2
6. A truck and a car are moving with same kinetic en-ergy. They are brought to rest by the application of brakes which provide equal force.
(A) both will cover equal distance
(B) the truck will cover a greater distance (C) the car will cover a greater distance (D) none of the above.
7. Newton's third law of motion can be used to ex-plain:
(A) Why the passengers in a bus tend to fall back-ward when its starts suddenly
(B) Swimming of a man (C) Motion of a rocket (D) Both (B) and (C).
8. Two blocks are kept in contact with each other on a smooth surface. The force on the lighter block and its acceleration is : -6Kg 10N 4Kg (A) 6 N, 1 m/s2 (B) 4 N, 1 m/s2 (C) 10 N, 1 m/s2 (D) 0 N, 0 m/s2
9. When the momentum of a body increases by 100% its kinetic energy increases by :
(A) 20% (B) 40%
(C) 200% (D) 300%
10. A stationary ball weighing 0.25 kg acquires a speed of 10 m/s when hit by a hockey stick. The impulse imparted to the ball is :
(A) 0.25 N × s (B) 2.5 N × s (C) 2 N × s (D) 0.5 N × s
11. Which of the following class of forces is different from others ?
(A) Pulling of a cart
(B) Stretching of a coiled spring (C) Kicking of a football (D) Weight of the body
12. External forces are : (A) always balanced (B) never balanced
(C) may or may not be balanced (D) none of these
13. If a force is conservative : (A) Work is path independent (B) Work is path dependent
(C) Potential energy remains constant (D) none of these
14. A body is in translatory equilibrium if : (A) Resultant force on it is zero (B) It is at rest
(C) It is in uniform motion (D) All options are correct 15. When a body is stationary :
(A) There is no force acting on it
(B) The forces acting on it are not in contact with it (C) The combination of forces acting on it balance each other
(D) The body is in vacuum
16. A particle is in straight line motion with uniform velocity. A force is not required :
(A) To increase the speed (B) To decrease the speed (C) To keep the speed constant (D) To change the direction
17. When a constant force is applied to a body, it moves with uniform :
(A) Acceleration (B) Velocity (C) Speed (D) Momentum 18. An object will continue accelerating until :
(A) Resultant force on it begins to decrease (B) Its velocity changes direction
(C) The resultant force on it is zero
(D) The resultant force is at right angles to its direction of motion
19. W hen a force of constant magnitude always act perpendicular to the motion of a particle then : (A) Velocity is constant
(B) Acceleration is constant (C) K.E. is constant (D) None of these
20. A block of mass 2 kg is placed on the floor. The coefficient of static friction is 0.4. If a force of 2.8 N is applied on the block parallel to floor the force of friction between the block and floor (taking g = 10 m/s2) is :
(A) 2.8 N (B) 8 N
PAGE # 11 21. Torque is the cause of :
(A) Translatory motion (B) Rotatory motion (C) Oscillatory motion
(D) Combined translatory and rotatory motion
22. It becomes easier to open or close a door turning about its hinges if the force is applied at the :
(A) Two third of the door (B) Free edge of the door (C) Middle of the door (D) Point near the hinges
23. Which of the following is non-conservative force ? (A) Electrostatic force (B) Gravitational force (C) Viscous force (D) Spring force
24. A box weighing 20 kg is pushed along the floor at a constant speed by applying a horizontal force. If the coefficient of friction is 0.25, then force applied is : (g = 10 m/s2)
(A) 5N (B) 10 N
(C) 50 N (D) 200 N
25. When a bicycle is in motion the force of friction exerted by the ground on the two wheels is such that it acts : (A) In the backward direction on the front wheel and in the forward direction on the rear wheel
(B) In the forward direction on the front wheel and in the backward direction on the rear wheel
(C) In the backward direction on both front and the rear wheels
(D) In the forward direction on both the front and rear wheels
26. A body of mass 20 kg is kept initially at rest. A force of 80 N is applied on the body then the acceleration produced in the body is 3 m/s2, force of friction acting on the body is :
(A) 80 N (B) 12 N
(C) 20 N (D) zero
27. It is required to increase the velocity of a scooter of mass 80 kg from 5 ms–1 to 25 ms–1 in 2 s. The force required will be :
(A) 200 N (B) 600 N (C) 800 N (D) 100 N
28. A car of mass 1000 kg is moving with a velocity of 10 m/s and is acted upon by a forward force of 1000 N due to engine and retarding force of 500 N. The velocity after 10 seconds will be :
(A) 10 m/s (B) 15 m/s (C) 20 m/s (D) zero
12
121212 PAGE # 12
STRUCTURE OF ATOM
DALTON’S ATOMIC THEORY
In 1808 John Dalton proposed atomic theory of matter, assuming atoms are ultimate indivisible particles of matter based on the law of conservation of mass and law of definite proportion.
The important points of Dalton’s theory are
-(i) Elements consist of small indivisible particles called atoms and atoms take part in chemical reactions.
(ii) Atoms of same element are alike in all respect.
(iii) Atoms of different elements are different in all respect.
(iv) Atoms cannot be created or destroyed.
(v) Atoms combine in a fixed, small, whole number to form compound atoms called molecules.
Note :
The term “ Element “ was coined by Lavoisier.
(a ) Me rits :
(i) Dalton’s theory explains the law of conservation of mass (point iv) and law of constant proportion (point v). (ii) Atoms of elements take part in chemical reaction this is true till today.
( b) D e me rit s :
(i) The atom is no longer supposed to be indivisible. The atom is not a simple particle but a complex one. (ii) He could not explain that why do atoms of same element combined with each other.
(iii) Atoms of the same element may not necessarily be identical in all aspects.
There are a number of elements whose atoms possess different masses. All these atoms of the same element with same atomic number but different mass number are called isotopes.
e.g. 1H1, 1H
2, 1H
3 are the three isotopes of hydrogen. (iv) Atoms of different elements may not necessarily be different in all aspects. There are a number of elements whose atoms possess same mass number. All these elements with different atomic number but same mass number are called isobars. e.g.20Ca40 and
18Ar
40 are isobars of each other.
ELECTRON
Electrons are the fundamental particles of all substances.
(a) Cathode Ray s - Discovery of Electron :
The nature and existence of electron was established by experiments on conduction of electricity through gases.
Note :
In 1859, Julius Plucker started the study of conduction of electricity through gases at low pressure in a discharge tube.
A number of interesting things happen when a high voltage (say, 10,000 V) is applied across the electrodes of the discharge tube, and the pressure of the gas inside the tube is lowered.
(i) When the pressure of the gas in the discharge tube is at atmospheric pressure and a high voltage is applied across the electrodes, nothing noticeable happens. But as we lower the pressure and increase the voltage, sparking or irregular streaks of light are seen in the tube. This is called positive column. (ii) As the pressure of gas is reduced further, the length of the positive column reduces, a fine glow can be seen at the cathode. The dark space or gap left between the cathode and the positive column is called the Faraday’s dark space.
13
131313 PAGE # 13 (iii) When the pressure of gas is reduced to about 1
mm of Hg, the cathode glow moves away from the cathode, creating a dark space between cathode and the cathode glow. This dark space is called Crookes dark space.
(iv) The Crookes dark space expands with further fall in pressure at 0.1 mm of Hg. The positive column gets split into a number of bands called striations. (v) At pressure 0.01 mm of Hg or less, the striations move towards the anode and vanish finally. At this stage the glass tubes begins to glow at the end opposite to the cathode. This phenomenon is called fluorescence.
Thus, some sort of invisible rays travel from the negative electrode to the positive electrode. Since the negative electrode is called cathode, these rays were called cathode rays. The colour of glow depends upon the nature of the glass used. For soda glass the fluorescence is of yellowish green colour.
( b) Propert ies of ca t hode ra y s :
(i) Cathode rays travel in a straight line at a high velocity and generate normally from the surface of the cathode. If an opaque object is placed in the path of cathode rays its shadow falls on opposite side of the cathode. It shows that cathode rays travel in straight lines. Cathode High voltage Anode To vacuum pump Shadow Object + – Note :
Cathode rays travel with very high velocities ranging from 109 to 1011 cm per second.
(ii) They are a beam of minute material particles having definite mass and velocity. When a light paddle wheel is placed in the path of the cathode rays, the blades of the paddle wheel begin to rotate. This also proves that cathode rays have mechanical energy.
(iii) They consist of negatively charged particles. When the cathode rays pass through an electric field, they bent towards the positive plate of the electric field. This indicates that cathode rays are negatively charged.
(iv) Cathode rays can affect the photographic plate. (v) The nature of cathode rays is independent of the nature of gas used in discharge tube or material of cathode.
(vi) Cathode rays are deflected in the magnetic field also. N S + – High voltage Deflection of cathode rays in magnetic field
(vii) If cathode rays are focused on a thin metal foil, the metal foil gets heated up to incandescence. (viii) When cathode rays fall on materials having high atomic mass, new type of penetrating rays of very small wavelength are emitted which are called X -rays.
Thus, investigations on cathode rays showed that these consisted of negatively charged particles. Note :
The negatively charged particles of cathode rays were called ‘negatrons‘ by Thomson. The name negatron was changed to ‘electron‘ by Stoney.
(c ) Charac te ristics of elec tron :
(i) Electrons are sub - atomic particles which constitute cathode rays.
(ii) In 1897, J.J.Thomson determined the charge to mass (e/m) ratio of electron by studying the deflections of cathode rays in electric and magnetic fields. The value of e/m has been found to be 1.7588 × 108 coulombs/g. The e/m for electrons from different gases was found to be the same. This indicates that atoms of all kinds have the same kind of negatively charged particles. Thus electrons are the common constituents of all atoms.
Note :
A cathode ray tube is used to measure the charge to mass ratio of the electrons.
14
141414 PAGE # 14 (iii) Charge on the electron :
The charge (e) on an electron was determined by Robert Millikan in 1909. Millikan found the charge on oil drops to be -1.6 × 10-19 Cor its multiples. So, the charge on an electron is to be -1.6 × 10-19 coulombs / unit. (iv) Mass of an electron :
By Thomson’s experiment e/m = 1.76 × 1011 C/kg By Millikan’s experiment e = – 1.6 × 10-19 C So mass of electron (m) = 11 19
10
1.76
10
1.6
= 9.1096 × 10-31 kg Mass of an electron in amu = 0.000549(v) Mass of electron in comparison to that of hydrogen :
Mass of hydrogen = 1.008 amu
= 1.008 × 1.66 × 10-24 g ( since 1 amu = 1.66 × 10-24 g ) = 1.673 × 10-24 g electron of Mass atom hydrogen of Mass = –28 24
10
9.1096
10
73
1.6
= 1837 Note :Thus, the mass of an electron is 1837
1
times the mass of a hydrogen atom.
PROTON
( a) Anode Ra ys (C anal ra ys ) :
It has been established that electron is a negatively charged particle and present in all the atoms. As an atom is electrically neutral, there must be some positively charged particles present in the atom to neutralize the negative charges of the electrons. It has been confirmed by experiments. Scientist Goldstein in 1886 discovered the existence of a new type of rays in the discharge tube. He carried out the experiment in discharge tube containing perforated cathode. It was observed that when high potential difference was applied between the electrodes, not only cathode rays were produced but also a new type of rays were produced simultaneously from anode, moving from anode towards cathode and passed through the holes of cathode.
Anode rays
Anode Perforated cathode
High voltage source
Fluorescence
Note :
Anode rays are called canal rays because they pass through the canals or holes of the cathode. These rays are also called anode rays since they originate from the anode side. Anode rays are produced from a positively charged electrode, therefore these were named positive rays by J.J.Thomson.
( b) C ha ra c t e ris t ic s of Anode Ra y s :
(i) Anode rays travel in straight lines.
(ii) These rays rotate the light paddle wheel placed in their path. This shows that anode rays are made up of material particles.
(iii) Anode rays are deflected by magnetic or electric field. In the electric field they get deflected towards negatively charged plate. This indicates that these rays are positively charged.
(iv) The anode rays affect photographic plate. (v) The nature of anode rays depend upon the type of gas used.
( c) D is c ov ery of Proton :
J.J.Thomson in 1906, found that particles obtained in the discharge tube containing hydrogen have e/m value as 9.579 × 104 coulomb/g. This was the maximum value of e/m observed for any positive particle. It was thus assumed that the positive particles given by hydrogen represent fundamental particle of positive charge. This particle was named proton.
H e H+ (Proton) Note :
The name ‘proton’ was given by Rutherford in 1911.
( d) Chara ct e rist ics of Prot on :
(i) A proton is a sub - atomic particle which constitute anode rays produced when hydrogen is taken in the discharge tube.
(ii) Charge of a proton :
Proton is a positively charged particle. The charge on a proton is equal but opposite to that on an electron. Thus, the charge on a proton is+1.602 × 10–19 coulombs/ unit.
(iii) Mass of a proton :
The mass of a proton is equal to the mass of a hydrogen atom.
mp = 1.0073 amu = 1.673 × 10-24 g = 1.673 × 10-27 kg
15
151515 PAGE # 15 (iv) Mass of proton relative to mass of electron :
electron an of Mass proton a of Mass = g 10 9.1 g 10 1.673 28 24 = 1837 Thus, the mass of a proton is 1837 times larger than the mass of an electron.
(v) Charge to mass ratio for a proton : The e/m of particles constituting the anode rays is different for different gases. m e of proton = 24 19 0 1 1.673 10 1.602 = 9.579 × 104 C/g
THOMSON MODEL OF AN ATOM
J.J. Thomson (1898) tried to explain the structure of atom. He proposed that an atom consists of a sphere of positive electricity in which electrons are embedded like plum in pudding or seeds evenly distributed in red spongy mass in case of a watermelon. The radius of the sphere is of the order 10–8 cm.
(a ) Me rits :
(i) Thomson’s model could explain the electrical neutrality of an atom.
(ii) Thomson’s mode l could explain why only negatively charged particles are emitted when a metal is heated as he considered the positive charge to be immovable by assuming it to be spread over the total volume of the atom.
(iii) He could explain the formation of ions and ionic compounds.
( b) D e me rit s :
This model could not satisfy the facts proposed by Rutherford through his alpha particle scattering experiment and hence was discarded.
RUTHERFORD MODEL OF AN ATOM
(a ) Rut he rford’s Alpha Pa rt ic le Sc at t e ring
Expe rime nt (1 9 0 9) :
Ernest Rutherford and his coworkers performed numerous experiments in which - particles emitted
from a radioactive element such as polonium were allowed to strike thin sheets of metals such as gold or platinum.
(i) A beam of -particles (He2+) was obtained by placing polonium in a lead box and letting the alpha particles come out of a pinhole in the lead box. This beam of
-rays was directed against a thin gold foil (0.0004 cm). A circular screen coated with zinc sulphide was placed on the other side of the foil.
(ii) About 99.0% of the -particles passed undeflected
through the gold foil and caused illumination of zinc sulphide screen.
(iii) Very few -particles underwent small and large
deflections after passing through the gold foil. (iv) A very few (about 1 in 20,000) were deflected backward on their path at an angle of 180º.
Rutherford was able to explain these observations as follows:
(i) Since a large number of -particles pass through
the atom undeflected, hence, there must be large empty space within the atom.
(ii) As some of the -particles got deflected, therefore,
there must be something massive and positively charged structure present in the atom.
(iii) The number of -particles which get deflected is
very small, therefore, the whole positive charge in the atom is concentrated in a very small space.
(iv) Some of the -particles retracted their path i.e.
came almost straight back towards the sources as a result of their direct collisions with the heavy mass.
Note :
- particles are made up of two protons and two neutrons and are Helium (He) nuclei.
(b) Rutherford Nuclear Model of Atom (1911) :
Rutherford proposed a new picture of the structure of atom.
Main features of this model are as follows-(i) The atom of an element consists of a small positively charged “Nucleus” which is situated at the centre of the atom and which carries almost the entire mass of the atom.
(ii) The electrons are distributed in the empty space of the atom around the nucleus in different concentric circular paths (orbits).
16
161616 PAGE # 16 (iii) The number of electrons in the orbits is equal to
the number of positive charges (protons) in the nucleus.
(iv) Volume of nucleus is very small as compared to the volume of atom.
(v) Most of the space in the atom is empty.
Note :
Rutherford’s model is also called “Planetary model’. ( c) D e fe c ts in Rut herford’s Mode l :
(i) Rutherford did not specify the number of electrons in each orbit.
(ii) According to electromagnetic theory, if a charged particle (like electron) is accelerated around another charged particle (like protons in nucleus) then there would be continuous loss of energy due to continuous emission of radiations. This loss of energy would slow down the speed of electron and eventually the electron would fall into the nucleus. But such a collapse does not occur. Rutherford’s model could not explain this theory.
(iii) If the electron loses energy continuously, the observed spectrum should be continuous but the actual observed spectrum consists of well defined lines of definite frequencies. Hence the loss of energy is not continuous in an atom.
BOHR MODEL OF AN ATOM (1913)
To overcome the objections to Rutherford’s model and to explain the hydrogen spectrum, Bohr proposed a quantum mechanical model of the atom.
The important postulates on which Bohr’s model is based are the following
-(i) The atom has a nucleus where all the protons are present. The size of the nucleus is very small. It is present at the centre of the atom.
(ii) Each stationary orbit is associated with a definite amount of energy. The greater is distance of the orbit from the nucleus, more shall be the energy associated with it. These orbits are also called energy levels and are numbered as 1, 2, 3, 4 ---or K, L, M, N ---- from nucleus to outwards.
(iii) By the time, the electron remains in any one of the allowed stationary orbits, it does not lose energy. Such a state is called ground or normal state. (iv) The emission or absorption of energy in the form of radiation can only occur when an electron jumps from one stationary orbit to another.
E = Efinal - Einitial = h
Where h is Planck’s constant (h = 6.625 × 10–34Js) Energy is absorbed when the electron jumps from lower to higher orbit and is emitted when it moves from higher to lower orbit.
When the electron moves from inner to outer orbit by absorbing definite amount of energy, the new state of the electron is said to be excited state.
(v) Negatively charged electrons revolves around the nucleus in circular path. The force of attraction between the nucleus and the electron is equal to centrifugal force of the moving electron.
Force of attraction towards nucleus = Centrifugal force (vi) Out of infinite number of possible circular orbits around the nucleus, the electron can revolve only in those orbits whose angular momentum is an integral multiple of 2 h , i.e. mvr = n 2 h where :
m = mass of the electron v = velocity of electron r = radius of the orbit, and n =1,2,3 ---- number of the orbit.
The angular momentum can have values such as
2 h , 2 h 2 , 2 h 3
, but it cannot have a fractional value. Thus, the angular momentum is quantized. The specified circular orbits (quantized) are called stationary orbits.
RADII OF VARIOUS ORBITS
Radii of various orbits can be given by formula.
r = 2 2 2 2 mkZe 4 h n Note :
Greater is the value of ‘n’ larger is the size of atom. On the other hand, greater is the value of ‘Z’ smaller is the size of the atom.
For hydrogen atom, Z = 1; so r = 2 2
2 2 mke 4 h n
Now putting the values of h, , m, e and k.
r = 2 –31 9 –19 2 2 34 – 2 ) 10 6 . 1 ( ) 10 9 ( ) 10 1 . 9 ( ) 14 . 3 ( 4 ) 10 625 . 6 ( n = 0.529 ×n2 × 10–10 m = 0.529 × n2 Å = 0.529 × 10–8 × n2 cm Thus, radius of 1st orbit
= 0.529 × 10–8 × 12 = 0.529 × 10–8 cm = 0.529 × 10–10 m = 0.529 Å Radius of 2nd orbit = 0.529 × 10–8 × 22 = 2.11 × 10–8 cm = 2.11 × 10–10 m = 2.11 Å Radius of 3rd orbit = 0.529 × 10–8 × 32 = 4.76 × 10–8 cm = 4.76 × 10–10 m = 4.76 Å
17
171717 PAGE # 17 Energy of an electron in Bohr’s orbit can be given
by the formula : E = 2 2 4 2 2 2 h n me K Z 2 –
For hydrogen atom, Z = 1 So, 2 2 4 2 2 h n me k 2 – E
Putting the values of , k, m, e and h.
E = – 34 2 – 2 4 19 – 31 – 2 9 2 ) 10 625 . 6 ( n ) 10 6 . 1 ( ) 10 1 . 9 ( ) 10 9 ( ) 14 . 3 ( 2 = – 2 19 – n 10 79 . 21 J per atom = 2 n 6 . 13
– eV per atom (1 J = 6.2419 × 1018 eV)
Note :
The negative sign indicates that the electron is under attraction towards nucleus, i.e. it is bound to the nucleus.
The electron has minimum energy in the first orbit and its energy increases as n increases, i.e., it becomes less negative. The electron can have a maximum energy value of zero when n = . The zero
energy means that the electron is no longer bound to the nucleus , i.e. , it is not under the force of attraction towards nucleus.
VELOCITY OF AN ELECTRON IN BOHR'S ORBIT Velocity of an electron in Bohr’s orbit can be given by the formula : v = n Z
h
e
2
2Substituting the values of h, , e.
v = n Z × 27 2 10 10 6.625 ) 10 4.8 ( 3.14 2 v = n Z × 2.188 × 108 cm/sec --- (iii) v = n 10 188 . 2 8
cm/sec (For hydrogen , Z = 1) v1 = 2.188 × 108 cm/sec v2 = 2 1 × 2.188 × 108 cm/sec = 1.094 × 108 cm/sec v3 = 3 1 × 2.188 × 108 cm/sec = 0.7293 × 108 cm/sec Here v1, v2 and v3 are the velocities of electron in first, second and third Bohr orbit in hydrogen.
NEUTRONS
In 1932, James Chadwick bombarded the element beryllium with - particles. He observed the emission
of a radiation with the following properties -(i) The radiation was highly penetrating.
(ii) The radiation remained unaffected in the electric or magnetic field i.e. the radiation was neutral. (iii) The particle constituting the radiation had the same mass as that of the proton. These neutral particles were called neutrons.
) (Beryllium Be 9 4 + particle) (á He 4 2 (Carbon) C 12 6 + (Neutron) n 1 0
COMPARATIVE STUDY OF ELECTRON,
PROTON AND NEUTRON
Property Electron Proton Neutron
Symbol e p n
Nature Negatively charged Positively charged Neutral Relative charge -1 +1 0 Absolute charge –1.602 × 10 -19 C +1.602 × 10-19 C 0 Relative mass 1 1 Absolute mass 9.109 × 10 -28 g 1.6725 × 10-24 g 1.6748 × 10-24 g 18 37 1 ATOMIC STRUCTURE An atom consists of two parts -(a) Nucleus
(b) Extra - nuclear region (a ) Nucleus :
Nucleus is situated at the centre of an atom. All the protons & neutrons are situated in the nucleus, therefore, the entire mass of an atom is almost concentrated in the nucleus. The overall charge of nucleus is positive due to the presence of positively charged protons (neutrons have no charge). The protons & neutrons are collectively called nucleons. Note :
The radius of the nucleus of an atom is of the order of 10–13 cm and its density is of the order of 1014 g/cm3. (b) Extra Nuc lear Re gion :
In extra nuclear part or in the region outside the nucleus, electrons are present which revolve around the nucleus in orbits of fixed energies. These orbits are called energy levels. These energy levels are designated as K, L, M, N & so on.
(i) The maximum number of electrons that can be accommodated in a shell is given by the formula 2n2.(n = number of shells i.e. 1,2,3 ---)
18
181818 PAGE # 18 Shell n 2n2 max. no.of electrons
K 1 2(1)2 2 L 2 2(2)2 8 M 3 2(3)2 18 N 4 2(4)2 32 K L 2 8 M N 18 32
+
Nucleus Electron shells
Maximum number of electrons which can be accommodated in the various shells First energy level
Second energy level Third energy level
Fourth energy level
(ii) Each energy level is further divided into subshells designated as s,p,d,f .
1st shell (K) contains 1 subshell (s) 2nd shell (L) contains 2 subshells (s,p) 3rd shell (M) contains 3 subshells (s,p,d) 4th shell (N) contains 4 subshells (s,p,d,f). (iii) Shells are divided into sub-shells, sub shells further contain orbitals.
(A) An orbital may be defined as
“A region in the three - dimensional space around the nucleus where the probability of finding the electron is maximum.”
(B)The maximum capacity of each orbital is that of two electrons.
Note :
The maximum number of orbitals that can be present in a shell is given by the formula n2.
(C) Types of orbitals :
(1) s-orbitals : The s-subshell contains just one orbital which is non-directional & spherically symmetrical in shape. The maximum number of electrons which can be accommodated in s-orbital is 2.
Z X Y
s- orbital
(2) p - orbitals : The p-subshell contains three orbitals which have dumb-bell shape and a directional character. The three p-orbitals are designated as px, py & pz which are oriented in the perpendicular axis (x,y,z). The maximum number of electrons which can be accommodated in the p subshell is 6 (2 electrons in each of three orbitals).
y y px z x y py z x pz y z x
(3) d - orbitals : The d-subshells contains 5 orbitals which are double dumb-bell in shape. These orbitals are designated as dxz, dxy, dyz,dx2 y2
,dz2. The d-subshell
can accommodate a maximum of 10 electrons.
y x dxy z y x dxz z y x dyz z z 2 2 z dz2 x y y x dx – y
(4) f-orbitals : The f-subshell contains 7 orbitals which are complex in structure.The f-subshell can accommodate a maximum of 14 electrons.
Note :
Letters s, p, d & f have originated from the words sharp, principle, diffused & fundamental respectively. (iv) Differences between orbit and orbital :
S.No. Orbit Orbital
1 It is well defined circular path around the
nucleus in which the electron revolves.
It is a region in three dimensional space around the nucleus where the probability of finding electron is maximum.
2 It is circular in shape. s,p and d-orbitals are spherical, dumb-bell and double dumb-bell in shape respectively.
3 It represents that an electron moves
around the nucleus in one plane.
It represents that an electron can move around nucleus along three dimensional space (x,y and z axis).
4
It represents that position as well as momentum of an electron can be known simultaneously with certainty. It is against Heisenberg's uncertainty principle.
It represents that position as well as momentum of an electron cannot be known simultaneously with certainty. It is in accordance with Heisenberg's uncertainty principle.
5
The maximum number of electrons in an orbit is 2n2 where 'n' is the number of the orbit.
The maximum number of electrons in an orbital is two.
19
191919 PAGE # 19 Note :
Heisenberg’s uncertainty principle - “It is impossible to determine exactly both the position and momentum (or velocity) of an electron or of any other moving particle at the same time.”
QUANTUM NUMBERS
To describe the position and energy of electron in an atom, four numbers are required, which are known as quantum numbers.
Four quantum numbers are : (a) Principal quantum number (b) Azimuthal quantum number (c) Magnetic quantum number (d) Spin quantum number
( a) Principal Qua nt um Numbe r : (i) It is denoted by ‘n’.
(ii) It represents the name, size and energy of the orbit or shell to which the electron belongs.
(iii) Higher is the value of ‘n’ , greater is the distance of the shell from the nucleus.
r1 < r2 < r3 < r4 < r5 <
----(iv) Higher is the value of ‘n’, greater is the magnitude of energy.
E1 < E2 < E3 < E4 < E5
----(v) Maximum number of electrons in a shell is given by 2n2.
Shell Max. number of electrons First (n =1) 2 × 12 = 2
Second (n = 2) 2 × 22 = 8 Third ( n = 3) 2 × 32 = 18 Fourth ( n = 4) 2 × 42 = 32
(vi) Angular momentum can also be calculated using principal quantum number.
2ð nh mvr
(vii) Value of n is from 1 to
(viii) Every shell is given a specific alphabetic name. First shell (n = 1) is known as K shell.
Second shell (n = 2) is known as L shell.
Third shell (n = 3) is known as M shell and so on. Note :
Principal quantum number was given by Bohr. ( b) Azimut ha l Qua nt um Numbe r : (i) It is represented by ‘’.
Note :
Azimuthal quantum number is also called angular quantum number, subsidiary quantum number or secondary quantum number.
(ii) For a given value of n values of is 0 to n – 1 Value of n Values of
1 (1st shell) 0 2 (2nd shell) 0,1 3 (3rd shell) 0,1,2 4 (4th shell) 0,1,2,3
(iii) It represents the sub-shell present in shell. = 0 represents s sub shell.
= 1 represents p sub shell. = 2 represents d sub shell. = 3 represents f sub shell.
Note :
s,p,d and f signify sharp, principal, diffused and fundamental respectively.
(iv) Number of sub-shell in a shell = Principal quantum number of shell.
(v) Maximum value of is always less than the value of n. So 1p, 1f, 2d, 2f, 3f subshells are not possible. s will start from 1s
p will start from 2p d will start from 3d f will start from 4f
(vi) Relative energy of various sub-shell in a shell are as follows
-s < p < d < f
(vii) Subshells having equal values but with different n values have similar shapes but their sizes increases as the value of ‘n’ increases. 2s-subshell is greater in size than 1s- subshell. Similarly 2p, 3p, 4p subshells have similar shapes but their sizes increase in order 2p < 3p < 4p.
(viii) Maximum no. of electrons present in a subshell = 2 (2 +1)
Subshell Max. electrons s ( = 0) 2 (2 × 0 +1) = 2 p ( = 1) 2 (2 × 1 +1) = 6
d ( = 2) 2 (2 × 2 +1) = 10 f ( = 3) 2 (2 × 3 +1) = 14 Note :
Azimuthal quantum number was given by Sommerfeld.
( c) Ma gne tic quant um numbe r : (i) It is denoted by ‘m’.
(ii) It represents the orbitals present in sub-shell. An orbital can be defined as :
“Region in the three - dimensional space around the nucleus where the probability of finding an electron is maximum”.
(iii) For a given value of , values of m are from –
through 0 to +. m 0 0 1 –1, 0, +1 2 –2, –1, 0, +1, +2 3 –3, –2, –1, 0, +1, +2, +3 (iv) Maximum number of orbitals in a sub-shell = (2+1)
Sub shell Orbitals s ( = 0) (2 × 0 +1) = 1 p ( = 1) (2 × 1 +1) = 3 d ( = 2) (2 × 2 +1) = 5