QJEEAdv12Chcb8f: JEE Advanced 2012 Chemistry - Class 11 JEEAdv12Chcb8f Solution: The kinetic energy of an electron in the second Bohr orbit of a …

For Bohr orbit, angular momentum is

$mvr_n = \frac{nh}{2\pi}$

Velocity, $v = \frac{nh}{2\pi mr_n} \quad \text{... (i)}$

Kinetic energy, $K.E. = \frac{1}{2}mv^2 \quad \text{... (ii)}$

By putting the value of $v$ from (i) into (ii),

$K.E. = \frac{1}{2}m \times \frac{n^2h^2}{4\pi^2m^2r_n^2} = \frac{n^2h^2}{8\pi^2mr_n^2}$

For second Bohr orbit, $n = 2$

$r_n = a_0 \times n^2 \quad (a_0 = \text{Bohr radius})$

$r_n = 4a_0$

$K.E. = \frac{(2)^2h^2}{8\pi^2m(4a_0)^2}$

Thus, $K.E. = \frac{h^2}{32\pi^2ma_0^2}$