Question

A free hydrogen atom after absorbing a photon of wavelength ${\lambda }_a$ gets excited from the state $n=1$ to the state $n=4$. Immediately after that the electron jumps to n = m state by emitting a photon of wavelength ${\lambda }_e$. Let the change in momentum of atom due to the absorption and the emission are $\Delta P_a$ and $\Delta P_e$ respectively. If ${\lambda }_a/{\lambda }_e=1/5$, which of the option (s) is/are correct ?
[Use $hc=1242eV$ nm ; $1nm={10}^{-9}m$, h and c are Plank's constant and speed of light, respectively]

• A. the ratio of kinetic energy of the electron in the state n = m to the state $n=1$ is $1/4$
• B. $m=2$
• C. $\Delta P_a/\Delta P_e=1/2$
• D. ${\lambda }_e=418nm$

Answer 1

Answer: (A), (B)

$m=2$

$\begin{array}{}\frac{hc}{{\lambda }_a}=13.6[\frac{1}{1}-\frac{1}{4^2}]& \text{(1)}\end{array}$
$\begin{array}{}\frac{hc}{{\lambda }_e}=13.6[\frac{1}{m^2}-\frac{1}{4^2}]& \text{(2)}\end{array}$
Divide equation $\left(2\right)$ by $\left(1\right)$
$\frac{{\lambda }_a}{{\lambda }_e}=\frac{\frac{1}{m^2}-\frac{1}{16}}{1-\frac{1}{16}}=\frac{1}{5}$
Since ${\lambda }_a/{\lambda }_e=1/5$
Solving we will get
$m=2$
Hence option B is correct
Since $\Delta P_a=\frac{h}{{\lambda }_a}$
$\Delta P_e=\frac{h}{{\lambda }_e}$
$\frac{\Delta P_a}{\Delta P_e}=\frac{{\lambda }_e}{{\lambda }_a}=5$
Hence option C is incorrect.
${\lambda }_e=\frac{1242\times 16}{13.6\times 3}\approx 487nm$
Hence option D is correct.
Since $K.E_n\propto \frac{z^2}{n^2}$
$\frac{K_2}{K_1}=\frac{1^2}{2^2}=\frac{1}{4}$
Hence option A is correct.