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 $K_2=\frac{1}{2^2K_1}=\frac{1}{4K_1}$
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