Question

The kinetic energy of an electron in the second Bohr orbit of a hydrogen atom is $\left[ a _{0}\right.$ is Bohr radius]

• A. $\frac{h^2}{4\pi^2 ma_0^2}$
• B. $\frac{h^{2}}{16 \pi^{2} m a_{0}^{2}}$
• C. $\frac{ h ^{2}}{32 \pi^{2} ma _{0}^{2}}$
• D. $\frac{h^{2}}{64 \pi^{2} ma _{0}^{2}}$

Answer 1

Answer: (C)

As per Bohr's postulate,
$mvr =\frac{ nh }{2 \pi}$
So, $v =\frac{ nh }{2 \pi mr }$
$KE =\frac{1}{2} mv ^{2}$
So, $KE =\frac{1}{2} m \left(\frac{ nh }{2 \pi mr }\right)^{2}$
Since, $r=\frac{a_{o} \times n^{2}}{z}$
So, for $2^{\text {nd }}$ Bohr orbit
$r=\frac{a_{0} \times 2^{2}}{1}=4 a_{a} $
$K E=\frac{1}{2} m\left(\frac{2^{2} h^{2}}{4 \pi^{2} m^{2} \times\left(4 a_{0}\right)^{2}}\right) $
$K E=\frac{h^{2}}{32 \pi^{2} m a_{0}^{2}}$