The kinetic energy of an electron in the second Bohr orbit of a h… - Vidyadip

MCQ

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 $10 \mathrm{x}$ is ........ . $\left(\mathrm{a}_{0}\right.$ is radius of Bohr's orbit) (Nearest integer) [Given : $\pi=3.14]$

A
$1010$
B
$6135$
✓
$3155$
D
$3845$

✓

Answer

Correct option: C.

$3155$

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

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

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

$\Rightarrow \mathrm{x}=315.507$

$\Rightarrow 10 \mathrm{x}=3155 \text { (nearest integer) }$

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