Question

Calculate the frequency of revolution of electron (in $10^{14}$ revolutions per sec) in the first Bohr orbit in a H-atom.

• A. 6.59
• B. 0.659
• C. 65.9
• D. 0.0659

Answer 1

Answer: (A)

65.9

Answer 2

The frequency of revolution ($f$) of an electron in the n-th Bohr orbit is given by $f = \dfrac{v}{2\pi r}$. For the first Bohr orbit ($n=1$) of a hydrogen atom ($Z=1$), the radius $r_1 \approx 5.29 \times 10^{-11}$ m and the speed $v_1 \approx 2.19 \times 10^6$ m/s.

$f_1 = \dfrac{2.19 \times 10^6 \text{ m/s}}{2\pi (5.29 \times 10^{-11} \text{ m})} \approx 6.59 \times 10^{15} \text{ Hz}$.

To express this in $10^{14}$ revolutions per second: $f_1 = 6.59 \times 10^{15} \text{ Hz} = 65.9 \times 10^{14} \text{ Hz}$.