Question

The number of revolutions per second made by an electron in the first Bohr's orbit of hydrogen atom is(Given, $r=0.53\stackrel{o}{A}$).

• A. $6\times {10}^5$Hz
• B. $2.5\times {10}^{12}$Hz
• C. $1.9\times {10}^{13}$Hz
• D. $6.5\times {10}^{15}$Hz

Answer 1

The correct option is D $6.5\times {10}^{15}$Hz

From Bohr's theory
$mvr=\frac{nh}{2\pi }$
For first orbit $n=1$
$m\left(r\omega \right)r=\frac{1\times h}{2\pi }$
$\Rightarrow m{r}^{2}2\pi f=\frac{h}{2\pi }$
$\Rightarrow f=\frac{h}{4{\pi }^{2}m{v}^2}$
$=\frac{6.6\times {10}^{-34}}{4\times (3.14{)}^2\times 9.1\times {10}^{-31}\times (0.33\times {10}^{-10}{)}^2}$.

=$6.5\times {10}^{15}Hz$