Bohr Model of Hydrogen Atom – An IIT-JEE 2007 question

The following MCQ appeared in IIT-JEE 2007 question paper:

The largest wave length in the ultraviolet region of the hydrogen spectrum is 122 nm. The smallest wave length in the infra red region of the hydrogen spectrum (to the nearest integer) is

(a) 802 nm (b) 823 nm (c) 1882 nm (d) 1648 nm

The wave number 1/λ, which is the number of waves per meter length in the case of hydrogen spectrum is given by Rydberg’s relation,

1/λ = R(1/n₁² – 1/n₂²) where R is Rydberg’s constant and n₁ and n₂ are integers.

Ultraviolet radiations are obtained in the Lyman series of hydrogen spectrum when electron transitions take place from higher orbits (of quantum number n>1)to the innermost orbit (of quantum number n=1). So, for the Lyman series, n₁=1 and n₂ = 2,3,4,…etc. The largest wave length in the Lyman series is obtained when the transition is from 2^nd orbit (n₂=2) to the first orbit (n₁=1).

The smallest wave length in the infra red region is obtained when electron transition occurs from the outermost orbit (n₂ = ∞) to the third orbit (n₁ = 3) and this spectral line is the shortest wave length line in the Paschen series ( for which n₁ = 3 and n₂ = 4,5,6….etc.).

For the largest wave length ultraviolet line we have (expressing the wave length in nanometer),

1/122 = R(1/1² – 1/2²) = 3R/4

For the smallest wave length(λ') infrared line we have

1/λ' = R(1/3² – 1/∞) = R/9

Dividing the first equation by the second, λ'/122 = 3×9/4, from which λ' = 823 nm.

Let us consider another similar question which appeared in Kerala Medical Entrance 2001 question paper:

Given that the longest wave length in Lyman series is 1240 Ǻ, the highest frequency emitted in Balmer series is

(a) 8×10¹⁴ Hz (b) 8×10¹² Hz (c) 8×10¹⁰ Hz (d) 8×10³ Hz (e) 8×10² Hz

In the Rydberg’s relation, 1/λ = R(1/n₁² – 1/n₂²), n₁=1 and n₂ = 2,3,4,…etc., for the Lyman series. For the Balmer series (which is in the visible region), n₁=2 and n₂ = 3,4,5….etc.

The longest wave length in Lyman series is obtained for n₂ = 2 and the highest frequency (shortest wavelength) in Balmer series is obtained for n₂ = ∞.

Rydberg’s relation for the above two cases are (expressing wave lengths in Angstrom),

1/1240 = R(1/1² – 1/2²) = 3R/4 and

1/ λ' = R(1/2² – 1/∞) = R/4

Dividing the first equation by the second,

λ'/1240 = 3, from which λ' = 3720 Ǻ.

The frequency (ν) of this line is given by

ν = c/λ' = (3×10⁸) /(3720×10^–10) = 8×10¹⁴ Hz.