Question

An orbital electron in a particular state of hydrogen has the radius $R$. This orbital electron is excited to another state by some energy transfer to the hydrogen atom so that its radius is now $9R$. Then

• A. Its energy is now one ninth of the original energy
• B. Its velocity is now one third of the original velocity
• C. Its angular momentum is three times the original angular momentum
• D. It will come to the original state by emitting maximum of 2 photons.

Answer 1

Answer: (A), (B), (C)

The correct options are
A Its energy is now one ninth of the original energy
B Its velocity is now one third of the original velocity
C Its angular momentum is three times the original angular momentum

Radius of an orbit
$r=0.529\frac{n_0^2}{z}{A}^{\circ }=R....\left(i\right)$
Now its radius become $9R$ then
$9R=0.529\frac{n^2}{z}....\left(ii\right)$
from (i) and (ii)
$n=3n_0$
energy of an $e^{-}$
$E=-13.6\frac{z^2}{n^2}$
$z=1$
$E=-\frac{13.6}{n^2}$
$\frac{E_n}{E_{n_0}}={\left(\frac{n_0}{n}\right)}^2$
$E_n=\frac{E{n}_0}{9}$
as angular - momentum
$L=\frac{nh}{2\pi }$
$\frac{L_n}{L_{n_0}}=3$
Now, $mvr=\frac{nh}{2\pi }$
$v\propto \frac{n}{r}$
$\frac{v_n}{v_{n_0}}=\left[\frac{n}{n_0}\right]\left[\frac{r_{n_0}}{r_n}\right]$
$\frac{V_n}{V_{n_0}}=\left[3\right]\left[\frac{R}{9R}\right]$
$\left[\frac{V_n}{V_{n_0}}\right]=\frac{1}{3}$
(D) only if initial state is ground state Hence, (A), (B), (C) are correct