CBSE Test Paper-02
Class - 12 Physics (Atoms)
1. According to Bohr's theory, the moment of momentum of an electron revolving in
second orbit of hydrogen atom will be:
a.
b.
c.
d.
2. The ground state energy of hydrogen atom is –13.6 eV. Find the orbital radius and
velocity of the electron in a hydrogen atom
a. 5.6 m, 2.5 m/s
b. 5.4 m, 2.3 m/s
c. 5.3 m, 2.2 m/s
d. 5.5 m, 2.4 m/s
3. A hydrogen atom is in a state with energy -1.51 eV. In the Bohr model, what is the
angular momentum of the electron in the atom, with respect to an axis at the nucleus?
a. 3.56
b. 3.16
c. 3.76
d. 3.36
4. The ground state energy of hydrogen atom is –13.6 eV. What are the kinetic and
potential energies of the electron in this state?
a. 13.6 eV, –27.2 eV
b. 14.6 eV, –27.2 eV
c. 14.6 eV, –29.2 eV
d. 13.1 eV, –29.2 eV
5. The ratio of longest wavelength and the shortest wavelength observed in the Balmer
series in the emission spectrum of hydrogen is:
a. 2.8
b. 1.8
c. 3.8
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d. 4.8
6. Name the series of hydrogen spectrum lying in the infrared region.
7. Can a hydrogen atom absorb a photon having energy more than 13.6 eV?
8. When is - line of the Balmer series in the emission spectrum of hydrogen atom
obtained?
9. The short wavelength limit for the Lyman series of the hydrogen spectrum is 913.4 A.
Calculate the short wavelength limit for Balmer series of the hydrogen spectrum.
10. Using Rutherford model of the atom, derive the expression for the total energy
of the electron in hydrogen atom. What is the significance of total negative
energy possessed by the electron?
11. A difference of 2.3 eV separates two energy levels in an atom. What is the frequency
of radiation emitted when the atom make a transition from the upper level to the
lower level?
12. Calculate the radius of the first orbit of hydrogen atom. Show that the velocity of
electron in the first orbit is times the velocity of light.
13. i. The energy levels of an atom are as shown in figure below. Which of them will
result in the transition of a photon of wavelength 275 nm?
ii. Which transition corresponds to emission of radiation of minimum wavelength?
14. The Rydberg constant for hydrogen is 10967700 m-1. Calculate the short and long
wavelength limits of Lyman series.
15. Determine the speed of electron in n = 3 orbit of He+. Is the non-relativistic
approximation valid?
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CBSE Test Paper-02
Class - 12 Physics (Atoms)
Answers
1. b.
Explanation: Angular momentum (L) is an integral multiple of where h
is the Planck's constant i.e. .
For second orbital electron, n=2, so
,
2. c. 5.3 m, 2.2 m/s
Explanation: Energy of electron,
E = -13.6 eV = -2.2 x 10 -18 J
Radius,
Velocity,
3. b.
Explanation: Energy in nth orbit = -13.6 / n2
n2 = -13.6/-1.51eV = 9
n=3
Angular momentum, L = nh/2pi
L = (3 6.626 10 -34) / 2 3.14
L = 3.16 10 -34 kgm2/sec
4. a. 13.6 eV, –27.2 eV
Explanation: Total energy, E = -13.6 eV
Since sum of energies is constant so
K.E = 13.6 eV (-E)
and P.E = -27.2 eV (2 K.E)
5. b. 1.8
Explanation: Longest wavelength in this series = 656.3 nm
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Shortest wavelenth = 364.6 nm
Ratio = 656.3 / 364.6 = 1.8
6. The series of lines in the hydrogen spectrum which lie in the infrared region are:
Paschen series : Near infrared region
Brackett series : Infrared region
Pfund series : Far infrared region
Humphrey series : Very far infrared region
7. Yes, it can absorb. But the atom would be ionized.
8. Line of the Balmer series in the emission spectrum of hydrogen atom is obtained
when an electron makes a transition from third lowest energy level to second lowest
energy level .i.e; From n=3 to n=2
9. Lyman series, n = 2, 3, 4... to n= 1
For short wavelength, n = to n = 1
we know that ,
Energy of nth orbit, E = 13.54/n2
So, energy of n = 1, energy level = 13.54eV
Energy of n = 2, energy level= 13.54/22 = 3.387 eV
So, short wavelength of Balmer series = = 3653
10. The Rutherford nuclear model of the atom describes the atom as an electrically
neutral sphere consisting of a very small, massive and positively charged nucleus at
the centre surrounded by the revolving electrons in their respective dynamically
stable orbits. The electrostatic force of attraction F, between the revolving electrons
and the nucleus provides the requisite centripetal force (Fc) to keep them in their
orbits. Thus, for a dynamically stable orbit in a hydrogen atom
Fc = Fe
[ Z = 1]
Thus, the relation between the orbit radius and the electron velocity is
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The kinetic energy (K) and electrostatic potential energy (U) of the electron in
hydrogen atom are
and
(The negative sign in U signifies that the electrostatic force is attractive in nature.)
Thus, the total mechanical energy E of the electron in a hydrogen atom is
The total energy of the electron is negative. This implies the fact that the electron is
bound to the nucleus. If E were positive, an electron will not follow a dosed orbit
around the nucleus and it would leave the atom.
11. E 2 -E 1= 2.3 eV
or
12. Since,
Using n = 1 for 1 st orbit
Z = 1 for hydrogen, coulomb
We get,
Also,
13. i. Given, wavelength of the photon,
Energy of photon is given by,
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From fig. this transition corresponds to B since for transition B.
E = 0 - (- 4.5eV) = 4.5eV
ii. Energy of Photon Emitted, E = hc/λ ∝ 1/ λ
For minimum wavelength of emission, the energy is maximum.
Transition D, for which the energy emission is maximum, corresponds to the
emission of radiation of minimum wavelength.
14. For Lyman series, the wave number is given by
For the short wavelength limit
or
For long wavelength limit
15. The speed of electron in nth orbit is given by
For He, Z = 2, n = 3
Now, = 0.048
which is much less than 1.
Hence non-relativistic approximation is true.
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