MHT CET · Physics · Atomic 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 \(\frac{\mathrm{m}^3 \mathrm{r}^3}{\mathrm{~h}^2}\)
- B \(\frac{\mathrm{h}^2}{\mathrm{~m}^2 \mathrm{r}^3}\)
- C \(\frac{\mathrm{h}^2}{\mathrm{mr}^3}\)
- D \(\frac{\mathrm{mr}^3}{\mathrm{~h}^2}\)
Answer & Solution
Correct Answer
(B) \(\frac{\mathrm{h}^2}{\mathrm{~m}^2 \mathrm{r}^3}\)
Step-by-step Solution
Detailed explanation
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 \(\mathrm{a}_{\mathrm{c}}=\frac{\mathrm{v}^2}{\mathrm{r}}\),
\(\begin{array}{ll}
\therefore & a_c=\frac{h^2}{4 \pi^2 m^2 r^3} \\
\therefore & a_c \propto \frac{h^2}{m^2 r^3}
\end{array}\)
\(\begin{aligned}
& \mathrm{mvr}=\frac{\mathrm{nh}}{2 \pi} \\
\therefore \quad v= & \frac{n h}{2 \pi \mathrm{mr}}=\frac{h}{2 \pi m r}(\because \mathrm{n}=1)
\end{aligned}\)
From centripetal acceleration \(\mathrm{a}_{\mathrm{c}}=\frac{\mathrm{v}^2}{\mathrm{r}}\),
\(\begin{array}{ll}
\therefore & a_c=\frac{h^2}{4 \pi^2 m^2 r^3} \\
\therefore & a_c \propto \frac{h^2}{m^2 r^3}
\end{array}\)
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