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(C) For a hydrogen atom,the radius of the $n^{th}$ orbit is given by $r = n^{2}a_{0}$.For the second orbit $(n=2)$,$r = 2^{2}a_{0} = 4a_{0}$.The kinet

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$3155$

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(C) For a hydrogen atom,the radius of the $n^{th}$ orbit is given by $r = n^{2}a_{0}$.For the second orbit $(n=2)$,$r = 2^{2}a_{0} = 4a_{0}$.The kinetic energy $(K.E.)$ of an electron is given by $K.E. = \frac{1}{2}mv^{2}$.From Bohr's postulate,$mvr = \frac{nh}{2\pi}$,so $v = \frac{nh}{2\pi mr}$.Substituting $v$ into the $K.E.$ expression: $K.E. = \frac{n^{2}h^{2}}{8\pi^{2}mr^{2}}$.Substituting $n=2$ and $r=4a_{0}$: $K.E. = \frac{2^{2}h^{2}}{8\pi^{2}m(4a_{0})^{2}} = \frac{4h^{2}}{8\pi^{2}m(16a_{0}^{2})} = \frac{h^{2}}{32\pi^{2}ma_{0}^{2}}$.Comparing this with $\frac{h^{2}}{xma_{0}^{2}}$,we get $x = 32\pi^{2}$.Given $\pi = 3.14$,$\pi^{2} = (3.14)^{2} = 9.8596$.$x = 32 	imes 9.8596 = 315.5072$.$10x = 3155.072$.Rounding to the nearest integer,$10x = 3155$.

The kinetic energy of an electron in the second Bohr orbit of a hydrogen atom is equal to $\frac{h^{2}}{xma_{0}^{2}}$. The value of $10x$ is ........ . ($a_{0}$ is radius of Bohr's orbit) (Nearest integer) [Given : $\pi=3.14$]

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