Q114 / 150 · MHT CET 2024 (15 May Shift 2) · Physics
Acceleration of an electron in the first Bohr's orbit is proportional to $\mathrm{m}=$ mass of electron, $\mathrm{r}=$ radius of the orbit, $\mathrm{h}=$ Planck's constant)
- A$\frac{\mathrm{m}^3 \mathrm{r}^3}{\mathrm{~h}^2}$
- Correct answerB$\frac{\mathrm{h}^2}{\mathrm{~m}^2 \mathrm{r}^3}$
- C$\frac{\mathrm{h}^2}{\mathrm{mr}^3}$
- D$\frac{\mathrm{mr}^3}{\mathrm{~h}^2}$
Solution
From Bohr's postulates,
$\begin{aligned}
& \mathrm{mvr}=\frac{\mathrm{nh}}{2 \pi} \\
\therefore \quad v= & \frac{n h}{2 \pi \mathrm{mr}}=\frac{h}{2 \pi m r}(\because \mathrm{n}=1)
\end{aligned}$
From centripetal acceleration $\mathrm{a}_{\mathrm{c}}=\frac{\mathrm{v}^2}{\mathrm{r}}$, $\begin{array}{ll} \therefore & a_c=\frac{h^2}{4 \pi^2 m^2 r^3} \\ \therefore & a_c \propto \frac{h^2}{m^2 r^3} \end{array}$
From centripetal acceleration $\mathrm{a}_{\mathrm{c}}=\frac{\mathrm{v}^2}{\mathrm{r}}$, $\begin{array}{ll} \therefore & a_c=\frac{h^2}{4 \pi^2 m^2 r^3} \\ \therefore & a_c \propto \frac{h^2}{m^2 r^3} \end{array}$
Hence, the correct answer is option: 'B'.