Subjective:
Board Type Questions
Problem 1:
Why are Bohr‟s orbits are called stationary states?
Solution:
This is because the energies of orbits in which the electrons revolve are fixed.
Problem 2:
Explain why the electronic configuration of Cu is 3d104s1 and not 3d94s2.
Solution:
In the 3d104s1 the d-subshell is completely filled which is more stable.
Problem 3:
Fe3+ ion is more stable than Fe2+ ion. Why?
Solution:
In Fe3+ ion 3d subshell is half filled hence more stable configuration.
Problem 4:
Calculate the accelerating potential that must be applied to a proton beam to give it an effective wavelength of 0.005 nm.
Solution:
v h m
1 2
eV mv
2
Putting the values we get V = 32.85 volt
Problem 5:
Give one example of isodiapheres.
Solution:
Isodiapheres have same difference between the number of neutrons and protons. For example
39 31
19 15
n p 1 n p 1
K & P
IIT Level Questions
On solving above equation n2 = 9
n = 3
Or corresponding transition from 3 2 in Balmer series of hydrogen atom has same frequency as that of 6 4 transition in He+.
Problem 7:
Calculate ionization potential in volts of (a) He and (b) Li
Calculate the ratio of K.E and P.E of an electron in an orbit?
Solution:
How many spectral lines are emitted by atomic hydrogen excited to
Thus the number of lines emitted from nth energy level = 1 + 2 + 3 +………… n – 1 = (n – 1)
Calculate (a) the de Broglie wavelength of an electron moving with a velocity of 5.0 105 ms–1 and (b) relative de Broglie wavelength of an atom of hydrogen and atom of oxygen moving with the same velocity
Solution:
(b) An atom of oxygen has approximately 16 times the mass of an atom of hydrogen. In the formula h
mv , h is constant while the conditions of problem make v, also constant. This means that and m are variables and varies inversely with m. Therefore, for the hydrogen atom would be 16 times greater than for oxygen atom.
Problem 11:
A 1 MeV proton is sent against a gold leaf (Z = 79). Calculate the distance of closest approach for head on collision.
Solution:
What is the wavelength associated with 150 eV electron
Solution:
Calculate the wavelengths of emitted radiation when the electron drops from third to second orbit.
Solution:
E3 – E2 = h = hc – 2.41 10–12 – (– 5.42 10–12) =
O2 undergoes photochemical dissociation into one normal oxygen and one excited oxygen atom, 1.967 eV more energetic than normal. The dissociation of O2 into two normal atoms of oxygen atoms requires 498kJ mole–1. What is the maximum wavelength effective for photochemical dissociation of O2?
Solution:
Energy required for excitation = 1.967 eV = 3.146 10–19J Total energy required for photochemical dissociation ofO2
= 8.268 10–19 + 3.146 10–19 = 11.414 10–19 J hc= 11.414 10–19J
= 6.626 10 3 1019
11.414 10 = 1.7415 10–7 m = 1741.5 Å
Problem 15:
Compare the wavelengths for the first three lines in the Balmer series with those which arise from similar transition in Be3+ ion. (Neglect
16times the hydrogen wavelengths.
Objective:
Problem 1:
For a p-electron, orbital angular moment is
(A) 2 (B)
(C) 6 (D) 2
Solution:
Orbital angular momentum L = ( 1)where h
2
L for p electron = 1(1 1) 2
(A)
Problem 2:
For which of the following species, Bohr theory doesn‟t apply
(A) H (B) He+
(C) Li2+ (D) Na+
Solution:
Bohr theory is not applicable to multi electron species (D)
Problem 3:
If the radius of 2nd Bohr orbit of hydrogen atom is r2. The radius of third Bohr orbit will be
(A) 4r2
Solution:
Circumference of 3rd orbit = 2 r3
According to Bohr‟s angular momentum of electron in 3rd orbit is mvr3 = 3 h
2 or h 2 r3
mv 3
By de-Broglie equation, = h
mv
= 2 r3
3
2 r3 = 3
i.e. circumference of 3rd orbit is three times the wavelength of electron or number of waves made by Bohr electron in one complete revolution in 3rd orbit is three.
(B)
Problem 5:
The degeneracy of the level of hydrogen atom that has energy -RH 16 is
(A) 16 (B) 4
(C) 2 (D) 1
Solution:
En = RH2
An electron is moving with a kinetic energy of 4.55 10–25 J. What will be de Broglie wavelength for this electron?
(A) 5.28 10–7 m (B) 7.28 10–7 m
(B)
Problem 7:
Suppose 10–17J of energy is needed by the interior of human eye to see an object. How many photons of green light ( = 550 nm) are needed to generate this minimum amount of energy?
(A) 14 (B) 28
(C) 39 (D) 42
Solution:
Let the number of photons required = n n hc 1017
n = 1017
hc = 1017 550 1034 9 8
6.626 10 3 10 = 27.6 = 28 photons (B)
Problem 8:
The two electrons present in an orbital are distinguished by
(A) principal quantum number (B) azimuthal quantum number (C) magnetic quantum number (D) spin quantum number
Solution:
They are distinguished by their spin.
(D)
Problem 9:
The velocity of electron in the ground state hydrogen atom is 2.18 106 ms–1. Its velocity in the second orbit would be
(A) 1.09 106 ms–1 (B) 4.38 106 ms–1 (C) 5.5 105 ms–1 (D) 8.76 106 ms
–1-Solution:
We know that velocity of electron in nth Bohr‟s orbit is given by = 2.18 106 Z
n m/s For H, Z = 1
v1 = 2.18 106
1 m/s
v2 = 2.18 106
2 m/s = 1.09 106 m/s (A)
Problem 10:
The ionization energy of the ground state hydrogen atom is 2.18 10–18J. The energy of an electron in its second orbit would be (A) –1.09 10–18 J (B) –2.18 10–18J
(C) –4.36 10–18J (D) –5.45 10–19J
Solution:
Energy of electron in first Bohr‟s orbit of H–atom E = 2.18 102 18J
n ( ionization energy of H = 2.18 10–18J) E2 = 2.18 102 18
2 J = –5.45 10–19J (D)
True and False
Problem 11:
All electromagnetic radiations have same energy.
Solution:
False.
Different electromagnetic radiation have different wavelengths, hence different energy
Problem 12:
Bohr‟s model is applicable for H-atom only.
Solution:
False.
Bohr‟s model is applicable for species having one electron like H, He+, Li2+.
Problem 13:
Specific charge of cathode rays remains same irrespective of the gas used in the discharge tube.
Solution:
True.
Cathode rays contain electrons which remains same with every gas.
Problem 14:
Isoelectronic species have same electronic configuration.
Solution:
True.
Isoelectronic species have same number of electrons.
Problem 15:
Atoms with same atomic number but different number of neutrons are called isotopes.
Solution:
True.
Isotopes have same atomic number but different atomic mass.
Fill in the Blanks
Problem 16:
The mass of positron is ___________ electron.
Solution:
Equal to
Problem 17:
With increasing principle quantum number, the energy difference between adjacent energy levels in H-atom _________.
Solution:
Decreases
Problem 18:
If the magnetic moment of an ion An+ is 5.9, then the number of unpaired electrons present in An+ are ____________.
Solution:
Five. Applying = n n 2 , where n 5
Problem 19:
s-electrons are more penetrating than p-electrons. The energy required to abstract s-electrons is ____________ than for p electrons.
Solution:
Greater
Problem 20:
The wavelength of the first spectral line in Paschen series for H-atom is____________.
Solution:
16410 Å
For Paschen series n2 4, n1 3