The kinetic energy of an electron in the second Bohr orbit of a hydrogen atom
Question:

The kinetic energy of an electron in the second Bohr orbit of a hydrogen atom is equal to $\frac{\mathrm{h}^{2}}{\mathrm{xma}_{0}^{2}}$. The value of $10 x$ is______________ ( $\mathrm{a}_{0}$ is radius of Bohr’s orbit)

(Nearest integer) [Given : $\pi=3.14]$

Solution:

$\mathrm{mvr}=\frac{\mathrm{nh}}{2 \pi}$

K.E. $=\frac{\mathrm{n}^{2} \mathrm{~h}^{2}}{8 \pi^{2} \mathrm{mr}^{2}}=\frac{4 \mathrm{~h}^{2}}{8 \pi^{2} \mathrm{~m}\left(4 \mathrm{a}_{0}\right)^{2}}$

$=\left(\frac{4}{8 \pi^{2} \times 16}\right) \frac{\mathrm{h}^{2}}{\mathrm{ma}_{0}^{2}}$

$\Rightarrow x=315.507$

$\Rightarrow 10 x=3155$ (nearest integer)

Administrator

Leave a comment

Please enter comment.
Please enter your name.