Atoms
1) Choose the correct alternative from the clues given at the end of each statement:
(a) The size of the atom in Thomson’s model is ---the atomic size in Rutherford’s model. (muc greater than/no different from/much less than) (b) In the ground state of ---electrons are in stable equilibrium, while in -- ---, electrons always experience a net force. (Thomson’s ,model/Rutherford’s model)
(c) A classical atom based on ---is doomed to collapse.
(Thomson’s model/Rutherford’s model) Solution:
(a) The size of the atom in Thomson’s model is no different from the atomic size in Rutherford’s model.
(b) In the ground state of Thomson’s model, electrons are in stable equilibrium, while in Rutherford’s model, electrons always experience a net force.
(c) A classical atom based on Rutherford’s model is doomed to collapse.
2) Choose the correct alternative from the clues given at the end of each statement
(d) An atom has a nearly continuous mass distribution in---, but has a highly non uniform mass distribution in Rutherford’s model.
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-.(Rutherford’s model/ Thomson’s model/both the models).
Solution:
(d) An atom has a nearly continuous mass distribution in Thomson’s model, but has a highly non uniform mass distribution in Rutherford’s model.
(e) The positivity charged part of the atom possesses most of the mass in both the models.
3) Suppose you are given a chance to repeat the alpha particle scattering experiment using a thin sheet of solid hydrogen in place of a gold foil.
(Hydrogen is a solid at temperatures below 14 K). What results do you expect?
Solution:
The basic purpose of the scattering experiment will be defeated, because solid hydrogen will be much lighter target compared to the alpha particle acting as the projectile. According to the theory of elastic collisions, the target hydrogen atoms will move with a momentum four times (much faster) that of the alpha-particle, after the collision. Hence, we will not be able to determine the size of the hydrogen nucleus.
4) What is the shortest wavelength present in the Paschen series of spectral lines?
Solution:
From Rydberg’s formula, = R [ − ]
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= 1.097 x 107 [ − ]
For the shortest wavelength in Paschen series, n1 = 3 and n2 = ∞ = 1.097 x 107 [ - ]
= 1.22 x 106 Λ = 8.2 x 10-7 m
= 820 nm
5) A difference of 2.3 e V separates two energy levels in an atom. What is the frequency of the radiation emitted when the atom makes a transition from the upper level to the lower level?
Solution:
According to the problem,
Energy of the radiation emitted, E = 2.3 e V = 2.3 x 1.6 x 10-19 J As, E = hv
Frequency of the radiation emitted, v = E/h
= . × . ×
. ×
= 5.6 x 1014 Hz
6) The ground state energy of the hydrogen atom is -13.6 eV. What are the kinetic and potential energies of the electron in this state?
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Ground state energy of hydrogen atom, E = -13.6 e V Kinetic energy of electron, KE = - E = 13.6 e V
Potential energy of the electron, PE=2E = -2 x 13.6 = -27.2 e V
7) A hydrogen atom initially in the ground level absorbs a photon, which excites it to the n = 4 level. Determine the wavelength and frequency of the photon.
Solution:
For ground state, n1 = 1, and n2 = 4 Energy of photon absorbed, E = E2 – E1 E = . - ( . ) e V
E = 13.6 [ − ] E = 13.6 [ - ] E = 13.6 x 15/16 e V E = 12.75 e V
E = 2.04 x 10-18 joule From, E = hc/λ
Wave length of the photon, λ = 12400/12.75 = 972.5 A0 Frequency of the photon, v = c/λ
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= ×
. ×
= 3.1 x 1015 Hz
8) (a) Using Bohr’s model, calculate the speed of the electron in a hydrogen atom in the n=1, 2 and 3 levels.
Solution:
(a) Velocity of electron in Bohr’s stationary orbit is. V =
= . × × m/s
Where, k = 1/4πε0 = 9 x 109 C2/Nm2,
‘n’ is the principal quantum number and ‘h’ is Planck’s constant.
For hydrogen, Z = 1
Speed of the electron in the first orbit (n=1), v1 = 2.18 x 106 m/s
Speed of the electron in the second orbit (n=2), v2 = 0.5 x 2.18 x 105 m/s = 1.09 x 106 m/s
Speed of the electron in the third orbit (n = 3), v3 = 1/3 x 2.18 x 105 m/s
= 7.3 x 105 m/s
9) (b) Using Bohr’s model, calculate the orbital period of the electron in a hydrogen atom in the n = 1, 2 and 3 levels.
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(b) Orbital period, T = 2πr/v
Radius of the first orbit of hydrogen, r1 = 0.53 x 10-10 m
Time period of revolution of the electron in the first orbit of hydrogen, T1 = × . ×
. ×
= 1.53 x 10-16 s
Radius of the orbit, r is directly proportional to n2, and velocity in the orbit, v is directly proportional to 1/n, so the time period of revolution of the electron in the orbit of hydrogen, T is directly proportional to n3.
Time period of revolution of the electron in the second orbit of hydrogen, n=2 T2 = 8T1 = 8 x 1.53 x 10-16 s
= 1.22 x 10-15 s
Time period of revolution of the electron in the third orbit of hydrogen, n=3 T3 = 27 T1 = 27 x 1.52 x 10-16 s = 4.1 x 10-15 s
10) The radius of the innermost electron orbit of a hydrogen atom is 5.3 x 10-
11 m. what are the radii of the n=2 and n=3 orbits?
Solution:
As, r is directly proportional to n2
The radius of the second orbit of hydrogen atom (n=2), r2 = 4r1
r 2 = 4 x 5.3 x 10-11 m = 2.12 x 10-10m
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The radius of the third orbit of hydrogen atom (n=3), r3 = 9r1 r3 = 9x5.3 x10-11m = 4.77 x 10-10 m
11) A 12.5 e V electron beam is used to bombard gaseous hydrogen at room temperature. What series of wavelengths will be emitted?
Solution:
In the ground state, the energy of gaseous hydrogen at room temperature = -13.6 e V.
When it is bombarded with 12.5 e V electron beam, the total energy of the electron becomes -13.6 + 12.5 = -1.1 e V.
The electron would jump from n=3 to n=2 giving rise to Balmer series.
From Rydberg’s formula 1/λ = R [ − ] 1/λ = 1.097 x 107 [¼ - 1/9]
Λ = 6.56 x 10-7 m = 656 nm
Itb may also jump from n=3 to n=1 or n=2 to n=1, giving rise to Lyman series.
For transition from n=2 to n=1
!" = 1.097 x 107 [ 1/1 – 1/9]
λLα = 1.03 x 10-7 m = 103 nm
12) In accordance with Bohr’s model, find the quantum number that characterizes the earth’s revolution around the sun in an orbit of radius 1.5 x 1011 m with orbital speed 3 x 104 m/s. (mass of earth = 6.0 x 1024 kg)
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According to the problem, Radius of earth, r = 1.5 x 1011 m Orbital speed of earth, v = 3 x 104 m/s Mass of earth, M= 6.0 x 1024 kg
According to Bohr’s model, mvr = nh/2π n = 2πmvr/h
n = 2 x x . × × × × . ×
. ×
n = 2.57 x 1074, which is too large.
13) Answer the following questions, which help you understand the difference between Thomson’s model and Rutherford’s model better.
(a) Is the average angle of deflection of α particles by a thin gold foil predicted by Thomson’s model much less, about the same, or much greater than that predicted by Rutherford’s model?
(b) Is the probability of backward scattering (i.e the scattering of α particle at angles greater than 900) predicted by Thomson’s model much less, about the same, or much greater than that predicted by Rutherford’s model?
Solution:
(a) About the same. This is because we are talking of the same average angle of deflection.
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(b) Much less, because in Thomson’s model, there is no such massive central core called the nucleus, contrary to Rutherford model.
14) Answer following questions, which help you understand the difference between Thomson’s model and Rutherford’s model better.
(c) Keeping the other factors fixed, it is found experimentally that for small thickness t, the number of α- particles scattered at moderate angles is proportional to t. what clues does this linear dependence on t provide?
(d) In which model is it completely wrong to ignore multiple scattering for calculating the average angle of scattering of α particle by a thin foil?
Solution:
(c) This suggests that the scattering is predominantly due to a single collision, because the chance of a single collision increases with the number of target atoms, which increases linearly with the thickness of the foil.
(d) In Thomson’s model, the positive charge is uniformly distributed in the spherical atom. Therefore, a single collision causes very little deflection.
Therefore, the average scattering angle can be explained only by considering multiple scattering. Thus, it is wrong to ignore multiple scattering when it comes from a single collision. Therefore, multiple scattering may be ignored as a first approximation.
15) The gravitational attraction between the electron and proton in a hydrogen atom is weaker than the coulomb attraction, by a factor of about 10-
40. An alternative way of looking at this is to estimate the radius of the first
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gravitational attraction. You will find the answer interesting. Solution:
The radius of the first Bohr orbit of a hydrogen atom is r0 = # $
%&' ()
If we consider the atom bound by gravitational force (F = *(+()
, ). We should replace
# by (-./. )
In that case, the radius of the first Bohr orbit of a hydrogen atom would be given by
r0 = $
%
&'
*(+()
Putting the standard values, we get
r0 = 1
. ×
& 2
. × × . × 3×4 . × ) = 1.2 x 1029 meters
This is much greater than the estimated size of the entire universe.
16) Obtain an expression for the frequency of radiation emitted when a hydrogen atom de excites from level n to level (n-1). For large n, show that this frequency equals the classical frequency of revolution of the electron in the orbit.
Solution:
The frequency v, of the emitted radiation when a hydrogen atom de excites from level n to level (n-1) is given by, E = hv = E2 – E1
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v = (5 6 [
4 ) - ] v = (5 6 [ 4 )
4 ) ] v = (5 6 4 )
4 )
For large n, (2n-1) ≈ 29, ;9< 49 − 1) ≈ 9 v = (5 6 = (5 6
Putting α = >
5
We get, v = (5 x >
5
v = > ---(1)
In Bohr’s model, velocity of electron in nth orbit, v = nh/2πmt Radius of nth orbit, r =
>
Frequency of revolution of electron, v = v/2πr = nh/2πmr ( >
? )
v = >
Which is same as (1)
Hence, for large values of n, the classical frequency of revolution of electron in the nth orbit is the same as the frequency of radiation emitted when a hydrogen atom de excites level (n) to level (n-1).
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Then what determines the typical atomic size? Why is an atom not, said, a thousand times bigger than its typical size? The question had greatly puzzled Bohr before he arrived at his famous model of the atom that you studied about in the text. To simulate what he might have done before his discovery, let us play as follows with the basic constants of nature and see if we can get a quantity with the dimensions of length that are roughly equal to the known size of an atom (~10-10m)
(a) Construct a quantity with the dimensions of length from the fundamental constants e, me and c. determine its numerical value.
Solution:
(a) Using fundamental constant e, me and c, we construct a quantity that has the dimensions of length. This quantity is (
# ()5 ) Now,
# ()5 = 4 . × ) × ×
. × 4 × )
= 2.82 x 10-15 m
This is much smaller than the typical atomic size.
18) Classically, an electron can be in any orbit around the nucleus of an atom.
Then what determines the typical atomic size? Why is an atom not, said, a thousand times bigger than its typical size? The question had greatly puzzled Bohr before he arrived at his famous model of the atom that you studied about in the text. To simulate what he might have done before his discovery, let us play as follows with the basic constants of nature and see if we can get a quantity with the dimensions of length that are roughly equal to the known
~10
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(b) You will find the length obtained in (a) is many orders of magnitude smaller than atomic dimensions. Further, it involves c. however, the energies of atoms are mostly in non relativistic domain where c is not expected to play any role. This is what may have suggested Bohr to discard c and look for something else to get the right atomic size. Now, Planck’s constant h had already made its appearance elsewhere. Bohr’s great insight lay in recognizing that h, me, and e will yield the right atomic size. Construct a quantity with the dimension of length from h, me, and e and confirm that its numerical value has indeed the correct order of magnitude.
Solution:
(b) However, when we drop c and use h, me, and e to construct a quantity that has dimensions of length, the quantity we obtain is # $
%&' ()
Now, # $
%&' ()
= 4 . × / )
× ×4 . × )4 . × )
≈ 0.53 x 10-10 m
This is of the order of atomic sizes.
19) The total energy of an electron in the first excited state of the hydrogen atom is about – 3.4 e V.
(a) What is the kinetic energy of the electron in this state?
(b) What is the potential energy of the electron in this state?
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potential energy is changed?
Solution:
We know that kinetic energy of electron, = KZe2/2r and PE of electron = - KZe2/r PE = -2 (kinetic energy) PE of electron and
In this calculation, the electric potential and hence the potential energy is zero at infinity.
Total energy = -3.4 e V
Total energy = PE + KE = -2KE + KE =-KE
(a) In the first excited state, kinetic energy = 3.4 e V
(b) PE of electron in this first excited state = -2 KE = -2 x 3.4 = -6.8 e V
(c) If the zero of potential energy is changed, the KE does not change and continuous to be +3.4 e V. however, the PE and total energy of the state would change with the choice of zero potential energy.
20) If Bohr’s quantization postulate (angular momentum = nh/2π) is a basic law of nature, it should be equally valid for the case of planetary motion also, why, then, do we never speak of the quantization of orbits of planets around the sun?
Solution:
Bohr’s quantization postulate is in terms of Plank’s constant (h). However, the angular momenta associated with planetary motion are ≈ 1070 h (for earth). In terms of Bohr’s quantization postulate, this will correspond to n≈ 1070. For such large values of n, the difference in successive energies and the angular momenta of
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the quantized levels are so small, that the levels can be considered as continuous and not discrete.
21) Obtain the first Bohr’s radius and the ground state energy of an muonic hydrogen atom [i.e. an atom in which a negatively charged muon (µ) of mass about 207 me orbits around a proton]
Solution:
Muonic hydrogen is an atom in which a negatively charged muon of mass about 207 me revolves around a proton
In Bohr’s atom model, r is directly proportional to 1/m
,E ,) = ()
(E ,E
,) = ()
() = 1/207
Here, re is radius of first orbit of electron in hydrogen atom = 0.53 A0 = 0.53 x 10-10 m
FG = ,) FG = . ×
FG = 2.56 x 10-13 m
Again, in Bohr’s atom model, E is directly proportional to m
HE H) = (E
(I
JG = 207 Ee
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Ee = -13.6 e V
JG = 207 (-13.6) e V
= -281502 e V = -2.8152 k e V
