Answer
$\frac{1}{4K_1}$
Work Step by Step
As we know that in the Bohr orbit, energy is directly proportional to $\frac{1}{n^2}$, so kinetic energy for $n=2$ in terms of $K_1$ is $K_2=\frac{1}{2^2K_1}=\frac{1}{4K_1}$
Physics Technology Update (4th Edition)
by Walker, James S.
Published by
Pearson
ISBN 10:
0-32190-308-0
ISBN 13:
978-0-32190-308-2
Chapter 31 - Atomic Physics - Problems and Conceptual Exercises - Page 1112: 14
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