MHT CET · Physics · Atomic Physics
In the Bohr model of hydrogen atom, the centripetal force is furnished by the coulomb attraction between the proton and the electron. If ' \(r_0\) ' is the radius of the ground state orbit, ' m ' is the mass, 'e' is the charge on the electron and ' \(\varepsilon_0\) ' is the permittivity of vacuum, the speed of the electron is
- A zero
- B \(\frac{\mathrm{e}}{\sqrt{\varepsilon_0 \mathrm{r}_0 \mathrm{~m}}}\)
- C \(\frac{\mathrm{e}}{\sqrt{4 \pi \varepsilon_0 \mathrm{r}_0 \mathrm{~m}}}\)
- D \(\frac{\sqrt{4 \pi \varepsilon_0 \mathrm{r}_0 \mathrm{~m}}}{\mathrm{e}}\)
Answer & Solution
Correct Answer
(C) \(\frac{\mathrm{e}}{\sqrt{4 \pi \varepsilon_0 \mathrm{r}_0 \mathrm{~m}}}\)
Step-by-step Solution
Detailed explanation
Given,
Centripetal Force \(=\) Coulomb attraction
\(\begin{aligned}
& \frac{\mathrm{mv}^2}{\mathrm{r}_0}=\frac{\mathrm{l}}{4 \pi \varepsilon_0} \frac{\mathrm{e}^2}{\mathrm{r}_0^2} \\
\therefore \quad & \mathrm{v}=\frac{\mathrm{e}}{\sqrt{4 \pi \varepsilon_0 \mathrm{r}_0 \mathrm{~m}}}
\end{aligned}\)
Centripetal Force \(=\) Coulomb attraction
\(\begin{aligned}
& \frac{\mathrm{mv}^2}{\mathrm{r}_0}=\frac{\mathrm{l}}{4 \pi \varepsilon_0} \frac{\mathrm{e}^2}{\mathrm{r}_0^2} \\
\therefore \quad & \mathrm{v}=\frac{\mathrm{e}}{\sqrt{4 \pi \varepsilon_0 \mathrm{r}_0 \mathrm{~m}}}
\end{aligned}\)
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