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Question: The electron in a hydrogen atom makes a transition n<sub>1</sub>® n<sub>2</sub>, where n<sub>1</sub>...

The electron in a hydrogen atom makes a transition n1® n2, where n1 and n2 are the principal quantum numbers of the two states. Assume Bohr model is valid in this case. The frequency of the orbital motion of the electron in the initial state is 1/27 of that in the final state. The possible values of n1 and n2 are-

A

n1= 6, n2 = 3

B

n1 = 4, n2 = 2

C

n1 = 8, n2 = 1

D

n1 = 3, n2 = 1

Answer

n1 = 3, n2 = 1

Explanation

Solution

f µ 1n3\frac{1}{n^{3}}

f1 = f227\frac{f_{2}}{27}

f1f2\frac{f_{1}}{f_{2}}= (n2n1)3\left( \frac{n_{2}}{n_{1}} \right)^{3}= 127\frac{1}{27}