Question

Which of the following transitions in the Bohr hydrogen atom results in the emission of the shortest wavelength photon.

a. n = 2 → n = 5

b. n = 5 → n = 2

c. n = 6 → n = 3

d. n = 3 → n = 6

e. n = 4 → n = 1

Answer 1

In the given transitions in the Bohr hydrogen atom results in the emission of the shortest wavelength photon use following expression:

rydberg formula
1 / λ= R * [ 1 / (n1 ^ 2) - 1 / (n2 ^ 2) ]
λ;lambda = wavelength in m
R = rydberg constant 10 973 731
n1 = inner shell
n2 = outer shell

λ= 1/ [ 1 / (n1 ^ 2) - 1 / (n2 ^ 2) ]

Lambda is indirectly proportional to [ 1/(n1^2) - 1/(n2^2) ] we can say that when the differecen is largest than they emitting shortest wavelength.

if electron transferred from outer shell to inner shell, photon is emitted, otherwise energy has to be supplyed

a. n = 2 → n = 5 thus here no proton produced

b. n = 5 → n = 2 ; [ 1/(n1^2) - 1/(n2^2) ] = [ 1/(2^2) - 1/(5^2) ]

= ¼-1-25= 0.25-0.04=0.21

c. n = 6 → n = 3; [ 1/(n1^2) - 1/(n2^2) ] = [ 1/(3^2) - 1/(6^2) ]

= 1/9-1-36= 0.11-0.027=0.083

d. n = 3 → n = 6 thus here no proton produced

e. n = 4 → n = 1; [ 1/(n1^2) - 1/(n2^2) ] = [ 1/(1^2) - 1/(4^2) ]

=1-1-16= 1.0-0.0625=0.09375

Thus here Lambda is indirectly proportional to [ 1/(n1^2) - 1/(n2^2) ] we can say that when the differecen is largest than they emitting shortest wavelength.

Here the largest difference is e. n = 4 → n = 1 in transition . Thus the correct answer is e.