The kinetic energy of an electron in the seco | vedclass

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(C) For a hydrogen atom,the radius of the $n^{th}$ Bohr orbit is given by $r_n = n^2 a_0$,where $a_0$ is the Bohr radius.For the second orbit $(n = 2)

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$\frac{h^2}{32\pi^2 m a_0^2}$

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(C) For a hydrogen atom,the radius of the $n^{th}$ Bohr orbit is given by $r_n = n^2 a_0$,where $a_0$ is the Bohr radius.For the second orbit $(n = 2)$,$r_2 = 2^2 a_0 = 4 a_0$.The velocity of an electron in the $n^{th}$ orbit is $v_n = \frac{nh}{2\pi m r_n}$.Substituting $r_n = n^2 a_0$,we get $v_n = \frac{nh}{2\pi m (n^2 a_0)} = \frac{h}{2\pi m n a_0}$.For $n = 2$,$v_2 = \frac{h}{2\pi m (2) a_0} = \frac{h}{4\pi m a_0}$.The kinetic energy $(KE)$ is given by $KE = \frac{1}{2} m v_2^2$.$KE = \frac{1}{2} m \left( \frac{h}{4\pi m a_0} \right)^2 = \frac{1}{2} m \left( \frac{h^2}{16\pi^2 m^2 a_0^2} \right) = \frac{h^2}{32\pi^2 m a_0^2}$.

The kinetic energy of an electron in the second Bohr orbit of a hydrogen atom is = ....... (Bohr radius $a_0$)

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