Q 12-12-124JEE MainJEE Main 2017 (9 Apr)Easy
The acceleration of an electron in the first orbit of the hydrogen atom $(n = 1)$ is:
Answer: (D) $\dfrac{h^2}{4\pi^2m^2r^3}$
Bohr's condition for $n = 1$: $mvr = \dfrac{h}{2\pi}$, so $v = \dfrac{h}{2\pi mr}$. The centripetal acceleration is
$$a = \frac{v^2}{r} = \frac{h^2}{4\pi^2m^2r^3}$$
Solution by Sreeraj P, M.Sc Physics