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 ?p and ?p , 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 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: (C)

m = 2

Answer 2

$\frac{hc}{\lambda_{a}} = 13.6\left[\frac{1}{1}-\frac{1}{4^{2}}\right]\quad...\left(i\right)$
$\frac{hc}{\lambda_{e}} = 13.6\left[\frac{1}{m^{2}}-\frac{1}{4^{2}}\right]\quad...\left(ii\right)$
(ii) / (i) , we get
$\frac{\lambda_{a}}{\lambda_{e}} = \frac{\left[\frac{1}{m^{2}}-\frac{1}{16}\right]}{\left[1-\frac{1}{16}\right]} = \frac{1}{5}$
$\Rightarrow\, \frac{1}{m^{2}}-\frac{1}{16} = \frac{15}{16}\times\frac{1}{5}$
$\Rightarrow\, \frac{1}{m^{2}}-\frac{1}{16} = \frac{3}{16}$
$\Rightarrow\, \frac{1}{m^{2}} = \frac{3}{16}+\frac{1}{16}$
$\Rightarrow\quad m = 2$
from (ii)
$\frac{hc}{\lambda_{e}} = 13.6\left[\frac{1}{2^{2}}-\frac{1}{4^{2}}\right] = 13.6\times\frac{3}{16}ev$
$\Rightarrow\quad\lambda_{e} = \frac{12400\times16}{13.6\times3}?$
$\Rightarrow\quad\lambda_{e} \approx 4862 ?$
we have $KE_{n} \,\,\propto \frac{z^{2}}{n^{2}}$
$\Rightarrow\, \frac{KE_{2}}{KE_{1}} = \frac{1}{4}$
$\Delta P_{a} = \frac{h}{\lambda_{a}}$
$\Delta P_{e} = \frac{h}{\lambda _{e}}$
$\Rightarrow\quad \frac{\Delta P_{a}}{\Delta P_{e}} = \frac{\lambda_{e}}{\lambda_{a}}$