The IIT-JEE Physics question paper is no more a nightmare to students aspiring for a seat in one of the IIT’s, consequent on the changes effected from 2007. If you have good grasp of fundamentals, you should definitely take up the challenge for brightening your future. Students appearing for other entrance examinations such as Physics GRE, AP Physics, AIEEE, KEAM and the like also will find the IIT-JEE question paper very useful for their preparation.

Here are three multiple choice questions on Bohr atom model which appeared under Linked Comprehension Type in IIT-JEE 2008 question paper:

**Paragraph for Question Nos. 1 to 3**

In a mixture of H – He

^{+}gas (He^{+}is singly ionized He atom) H atoms and He^{+}ions are excited to their respective first excited states. Subsequently, H atoms transfer their total excitation energy to He^{+}ions (by collisions). Assume that the Bohr model of atom is exactly valid.**(1)**The quantum number

*n*of the state finally populated in He

^{+}ions is

(A) 2

(B) 3

(C) 4

(D) 5

The singly ionised helium atom is hydrogen like since it contains a single electron. In a hydrogen like atom the energy (

*E*) of the electron in the n^{th}orbit is given by*E*= – 13.6 Z

^{2}/n

^{2}

*electron volt*where Z is the atomic number.

For hydrogen atom the energies are – 13.6 eV, – 3.4 eV, – 1.51 eV and – 0.85 eV for values of equal to 1, 2, 3 and 4 respectively

*n*
For the helium ion the energies are four fold, equal to – 54.4 eV, – 13.6 eV, – 6.04 eV and – 3.4 eV for values of

*n*equal to 1, 2, 3 and 4 respectively
The energy difference between the electrons in first excited states of hydrogen atom and helium ion is (– 3.4 eV) – (– 13.6 eV) = 10.2 eV.

On collision, the helium ion absorbs this energy and occupies the 3

^{rd}excited state (**)***n*= 4
of energy – 13.6 eV + 10.2 eV = –3.4 eV.

Therefore the correct option is (C).

**(2)**The wave length of light emitted in the visible region by He

^{+}ions after collision with H atoms is

(A) 6.5×10

^{–7}m
(B) 5.6×10

^{–7}m
(C) 4.8×10

^{–7}m
(D) 4.0×10

^{–7}m
The visible photons have wave lengths in the range 4000 Ǻ to 7000 Ǻ and
their energies lie in the range 3.1 eV to 1.771 eV.

[It will be useful to
remember the range of energies in eV. You will definitely remember the wave
length range of visible photons which you can convert into energy range using
the equation,

*λ E*= 12400 where*λ*is the wave length in Angstrom and*E*is the energy in electron volt].
The energy difference between the He

^{+}ion levels for*n*= 3 and*n*= 4 correspond to visible photon energy: – 3.4 eV– (– 6.04 eV) = 2.64 eV
The wave length of the light emitted in the visible region is therefore 124000/2.64
= 4800 Ǻ, nearly. The answer is thus 4.8×10

^{–7}m [Option (C)].**(3)**The ratio of the kinetic energy of the

*n*= 2 electron for the H atom to that of the He

^{+}ion is

(A) ¼

(B) ½

(C) 1

(D) 2

The kinetic energy of the electron in the n

^{th}orbit is 13.6 Z^{2}/n^{2}.
[Note that the kinetic energy and the total energy of the electron are equal in amount. But the kinetic energy is positive where as the total energy is negative].

The required ratio is Z

_{H}^{2}/Z_{He}^{2}= ¼.
All posts on Bohr model of hydrogen atom on this site can be accessed by clicking on the label ‘Bohr model’ below this post.

Thank you for the solutions. :)

ReplyDeletei was unable to understand before...but now it feels damn easy

ReplyDeletethank u so much sir

nice

ReplyDelete