A free hydrogen atom after absorbing a photon of wavelength _ a g… - Vidyadip

MCQ

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 $\lambda_{\text {. }}$. Let the change in momentum of atom due to the absorption and the emission are $\Delta p_2$, and $\Delta p_\theta$, respectively. If $\lambda_2 / \lambda_0=\frac{1}{5}$. 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]

$(1)$ $\lambda_0=418 nm$

$(2)$ The ratio of kinetic energy of the electron in the state $n = m$ to the state $n =1$ is $\frac{1}{4}$

$(3)$ $m =2$

$(4)$ $\Delta p_{ a } / \Delta p _{ o }=\frac{1}{2}$

✓
$2,3$
B
$2,4$
C
$3,2$
D
$1,3$

✓

Answer

Correct option: A.

$2,3$

a
$\frac{ hc }{\lambda_{ a }}=13.6\left[\frac{1}{1}-\frac{1}{4^2}\right]$ $. . . . . . .(1)$

$\frac{ hc }{\lambda_0}=13.6\left[\frac{1}{ m ^2}-\frac{1}{4^2}\right]$ $. . . . . . .(1)$

$(ii) / (i),$ we get

$\frac{\lambda_{ a }}{\lambda_0}=\frac{\left[\frac{1}{ m ^2}-\frac{1}{16}\right]}{\left[1-\frac{1}{16}\right]}=\frac{1}{5}$

$\Rightarrow \frac{1}{ m ^2}-\frac{1}{16}=\frac{15}{16} \times \frac{1}{5}$

$\Rightarrow \frac{1}{ m ^2}-\frac{1}{16}=\frac{3}{16}$

$\Rightarrow \frac{1}{ m ^2}=\frac{3}{16}+\frac{1}{16}$

$\Rightarrow m =2$

from $(ii)$

$\frac{ hc }{\lambda_0}=13.6\left[\frac{1}{2^2}-\frac{1}{4^2}\right]=13.6 \times \frac{3}{10} ev$

$\Rightarrow \lambda_0=\frac{12400 \times 16}{13.6 \times 3} \mathring A$

$\Rightarrow \lambda_0 \approx 4862 \mathring A$

we have $K E_u \propto \frac{z^2}{n^2}$

$\Rightarrow \frac{ KE _2}{ KE _1}=\frac{1}{4}$

$\Delta P _{ a }=\frac{ h }{\lambda_{ a }}$

$\Delta P _0=\frac{ h }{\lambda_0}$

$\Rightarrow \frac{\Delta P _{ a }}{\Delta P }=\frac{\lambda_0}{\lambda}$

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