Question

For an electron placed in the third orbit of the hydrogen atom, calculate the number of revolutions per second made by it around the nucleus.

• A. $4.2\times {10}^{14}$
• B. $4.2\times {10}^{12}$
• C. $2.4\times {10}^{12}$
• D. $2.4\times {10}^{14}$

Answer 1

The correct option is D $2.4\times {10}^{14}$

Radius of 3rd orbit $=3^2\times 0.529\times {10}^{-8}\text{cm}=4.761\times {10}^{-8}\text{cm}$
We know that,
$m\text{v}r=\frac{nh}{2\pi }$

$\Rightarrow \text{v}=\frac{nh}{2\pi mr}$

$\Rightarrow \text{v}=\frac{3\times 6.626\times {10}^{-27}}{2\times 3.14\times (9.1\times {10}^{-28})\times (4.761\times {10}^{-8})}$

$=7.3\times {10}^7\text{cm/sec}$

Number of revolutions per second $=\frac{\text{v}}{2\pi r}=\frac{7.3\times {10}^7}{2\times 3.14\times 4.761\times {10}^{-8}}$

$=2.4\times {10}^{14}$