Question

A free hydrogen atom after absorbing a photon of wavelength ${\lambda }_a$ gets excited from state $n=1$ to $n=4$. Immediately after electron jumps to $n=m$ state by emitting a photon of wavelength ${\lambda }_e$. Let change in momentum of atom due to the absorption and the emission are $\Delta P_a$ and $\Delta p_e$ respectively. If $\frac{{\lambda }_a}{{\lambda }_e}=\frac{1}{5}$. Which of the following is correct

• A. $m=2$
• B. $\frac{\Delta P_a}{P_e}=\frac{1}{2}$
• C. ${\lambda }_e=418nm$
• D. Ratio of K.E. of electron in the state$n=mton=1is\frac{1}{4}$

Answer 1

Answer: (A), (D)

Ratio of K.E. of electron in the state$n=mton=1is\frac{1}{4}$

Step 1. Given Data:

${\lambda }_a$ = Wavelength of a photon absorbed

${\lambda }_e$ = Wavelength of a photon emitted

$\Delta P_a$ = Change in momentum of atom due to the absorption

$\Delta p_e$ = Change in momentum of atom due to the emission

$\frac{{\lambda }_a}{{\lambda }_e}=\frac{1}{5}$

Step 2. As we know that:

$\frac{{\lambda }_a}{{\lambda }_e}=\frac{E_4-E_1}{E_4-E_m}$ [$E$ is the electric field]

$$\begin{gathered} \Rightarrow \frac{{\lambda }_a}{{\lambda }_e}=\frac{\left(1-{\displaystyle \frac{1}{16}}\right)}{\left({\displaystyle \frac{1}{m^2}-\frac{1}{16}}\right)} \\ \Rightarrow \frac{{\lambda }_a}{{\lambda }_e}=\frac{1}{5} \\ \therefore m=2 \end{gathered}$$

Step 3. Further:

$$\begin{gathered} {\lambda }_e=\frac{12400\times 4}{13.6} \\ \Rightarrow {\lambda }_e=3647 \\ \Rightarrow \frac{K_2}{K_1}=\frac{1^2}{2^2} \\ \therefore \frac{K_2}{K_1}=\frac{1}{4} \end{gathered}$$ [$K_2$ is the kinetic energy to $n=m$ and $K_1$is the kinetic energy to $n=1$]

Hence, the correct option are both (A) & (D).