Q121 / 180 · COMEDK 2025 (10 May Shift 2) · Physics
What is the frequency ' $v$ ' of the electron in Bohr's first orbit of radius ' $r$ ' of the hydrogen atom?
- A$v=\dfrac{e^2}{2 \pi \varepsilon_0 h r}$
- Correct answerB$v=\dfrac{e^2}{4 \pi \varepsilon_0 h r}$
- C$v=\dfrac{e^2}{2 \pi \varepsilon_0 h r^2}$
- D$v=\dfrac{e^2}{4 \pi \varepsilon_0 h r^2}$
Solution
For $n=1$, Bohr's quantization gives: $mv_{orb}r = \dfrac{h}{2\pi}$ ... (1)
Electrostatic force provides centripetal force: $\dfrac{mv_{orb}^2}{r} = \dfrac{e^2}{4\pi\epsilon_0 r^2}$ ... (2)
Dividing (2) by (1):
$\dfrac{v_{orb}}{r^2} = \dfrac{e^2 \cdot 2\pi}{4\pi\epsilon_0 r^2 h}$
$v_{orb} = \dfrac{e^2}{2\epsilon_0 h}$
Frequency of revolution: $v = \dfrac{v_{orb}}{2\pi r} = \dfrac{e^2}{2\epsilon_0 h \cdot 2\pi r} = \dfrac{e^2}{4\pi\epsilon_0 hr}$
Electrostatic force provides centripetal force: $\dfrac{mv_{orb}^2}{r} = \dfrac{e^2}{4\pi\epsilon_0 r^2}$ ... (2)
Dividing (2) by (1):
$\dfrac{v_{orb}}{r^2} = \dfrac{e^2 \cdot 2\pi}{4\pi\epsilon_0 r^2 h}$
$v_{orb} = \dfrac{e^2}{2\epsilon_0 h}$
Frequency of revolution: $v = \dfrac{v_{orb}}{2\pi r} = \dfrac{e^2}{2\epsilon_0 h \cdot 2\pi r} = \dfrac{e^2}{4\pi\epsilon_0 hr}$
Hence, the correct answer is option: 'B'.