Question

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

• A. $1.23\times {10}^{13}$
• B. $4.2\times {10}^{12}$
• C. $8.22\times {10}^{14}$
• D. $2.4\times {10}^{15}$

Answer 1

The correct option is C $8.22\times {10}^{14}$

We know, radius of $n^{th}$ orbit r = $0.529\times {10}^{-8}{n}^2$ cm
radius of $2^{nd}$ orbit $=2^2\times 0.529\times {10}^{-8}\text{cm}=2.12\times {10}^{-8}\text{cm}$
Also,
$m\text{v}r=\frac{nh}{2\pi }$

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

$\Rightarrow \text{v}=\frac{2\times 6.626\times {10}^{-27}}{2\times 3.14\times (9.1\times {10}^{-28})\times (2.12\times {10}^{-8})}$

$=1.09\times {10}^8\text{cm/sec}$

Number of revolutions per second $=\frac{\text{v}}{2\pi r}=\frac{1.09\times {10}^8}{2\times 3.14\times 2.12\times {10}^{-8}}$

$=8.22\times {10}^{14}$