Chemistry - Atomic Structure Question with Solution | TestHub

De Broglie's equation for calculating the wave length$\left(\mathrm{\lambda }\right)$ associated with the electron is given below

$\lambda =\frac{h}{p}$$=\frac{\mathrm{h}}{\mathrm{mV}}$

Let Kinetic energy of the particle of mass 'm' is K

$$\begin{gathered} \mathrm{K}=\frac{1}{2}{\mathrm{mV}}^2 \\ 2\mathrm{Km}={\mathrm{m}}^2{\mathrm{V}}^2 \\ \sqrt{2\mathrm{Km}}=\mathrm{mV} \end{gathered}$$

$\mathrm{\lambda }=\frac{\mathrm{h}}{\sqrt{2\mathrm{mK}}}$

$\mathrm{h}$=Planck's constant=$6.62\times {10}^{-27}\mathrm{erg}-\sec$

$\mathrm{K}=\frac{{\mathrm{h}}^2}{2 {\mathrm{m\lambda }}^2}$

$\mathrm{K}=\frac{{\mathrm{h}}^2}{2 {\mathrm{m\lambda }}^2}=\frac{43.9\times {10}^{-68}}{2\times 9.1\times {10}^{-31}\times 10.89\times {10}^{-20}}$

$\mathrm{K}=2.215\times {10}^{-18}$

$$\begin{gathered} {\mathrm{E}}_{\mathrm{abs}}={\mathrm{E}}_{\mathrm{req}}+\mathrm{K} \\ {\mathrm{E}}_{\mathrm{abs}}=\mathrm{Energy} \mathrm{absorbed} \\ {\mathrm{E}}_{\mathrm{req}=}\mathrm{Energy} \mathrm{required} \\ \mathrm{K}=\mathrm{Kinetic} \mathrm{energy} \end{gathered}$$$\frac{{\mathrm{E}}_{\mathrm{abs}}}{{\mathrm{E}}_{\mathrm{req}}}=1+\frac{\mathrm{K}}{{\mathrm{E}}_{\mathrm{req}}}=1+\frac{2.215\times {10}^{-18}}{13.6\times 1.602\times {10}^{-19}}=2.0166$