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Q.
The angular momentum of an electron in hydrogen atom is $\frac{h}{\pi }$ . The kinetic energy of the electron is
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Solution:
The angular momentum of an electron in hydrogen atom $=\frac{h}{\pi }$ .
As we know, from Bohr's second postulate,
$mvr=\frac{h}{\pi }\Rightarrow v=\frac{h}{\pi m r}$ ...(i)
Now, kinetic energy of electron
$=\frac{1}{2}mv^{2}=\frac{1}{2}m\times \left(\frac{h}{\pi m r}\right)^{2}$ [from Equation (i)]
$=\frac{h^{2}}{2 \pi ^{2} m r^{2}}$
$=\frac{\left(6.63 \times 10^{-34}\right)^2}{2 \times(3.14)^2 \times 9.1 \times 10^{-31} \times\left(5.3 \times 10^{-11}\right)^2 \times 1.6 \times 10^{-19}}$
$=3.4 \, \text{e}\text{V}$