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 ${\lambda }_a/{\lambda }_e=\frac{1}{5}$ . Which of the option(s) is/are correct ?
[Use hc = 1242 eV nm; 1 nm = $10^{-9}m$, h and c are Planck's constant and speed of light, respectively]

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

Answer 1

Answer: (B), (C)

$\frac{hc}{{\lambda }_a}=13.6\left[\frac{1}{1}-\frac{1}{4^2}\right]...\left(i\right)$
$\frac{hc}{{\lambda }_e}=13.6\left[\frac{1}{m^2}-\frac{1}{4^2}\right]...\left(ii\right)$
(ii) / (i) , we get
$\frac{{\lambda }_a}{{\lambda }_e}=\frac{\left[\frac{1}{m^2}-\frac{1}{16}\right]}{\left[1-\frac{1}{16}\right]}=\frac{1}{5}$
$\Rightarrow \frac{1}{m^2}-\frac{1}{16}=\frac{15}{16}\times \frac{1}{5}$
$\Rightarrow \frac{1}{m^2}-\frac{1}{16}=\frac{3}{16}$
$\Rightarrow \frac{1}{m^2}=\frac{3}{16}+\frac{1}{16}$
$\Rightarrow m=2$
from (ii)
$\frac{hc}{{\lambda }_e}=13.6\left[\frac{1}{2^2}-\frac{1}{4^2}\right]=13.6\times \frac{3}{16}ev$
$\Rightarrow {\lambda }_e=\frac{12400\times 16}{13.6\times 3}\mathring{A}$
$\Rightarrow {\lambda }_e\approx 4862\mathring{A}$
we have $K{E}_n\propto \frac{z^2}{n^2}$
$\Rightarrow \frac{K{E}_2}{K{E}_1}=\frac{1}{4}$
$\Delta P_a=\frac{h}{{\lambda }_a}$
$\Delta P_e=\frac{h}{{\lambda }_e}$
$\Rightarrow \frac{\Delta P_a}{\Delta P_e}=\frac{{\lambda }_e}{{\lambda }_a}$