Question

Of the following transitions in the Bohr hydrogen atom, the _ transition results in the emission of lowest-energy photon:

(A) n = 2 --> n = 1

(B) n = 6 --> n = 1

(C) n = 6 --> n = 5

(D) n = 5 --> n = 6

(E) n = 1 --> n = 6

Answer 1

The energy of emitted photon is equal to the energy difference between the initial orbit and final bohr's orbit.

A)

= -2.18 x 10^-18 x (3/4)

= - 1.635 x 10^-18 J

B)

= -2.18 x 10^-18 x (35/36)

= - 2.1194 x 10^-18 J

C)

= -2.18 x 10^-18 x (11/[25 x 36])

= -0.2664 x 10^-18 J

D)

= -2.18 x 10^-18 x (-11/[25 x 36]) =  

= 0.2664 x 10^-18 J

This involves absorption.

E)

= -2.18 x 10^-18 x (-35/36) = = 2.1194 x 10^-18 J

This involves absorption.

So the answer is C