Chemistry - Atomic Structure Question with Solution | TestHub

Energy incident $=\frac{\mathrm{hc}}{\lambda }$

$=\frac{6.63\times {10}^{-34}\times 3.0\times {10}^8}{248\times {10}^{-9}\times 1.6\times {10}^{-19}} \mathrm{eV}$

$=\frac{6.63\times 3\times 100}{248\times 1.6}$

$=0.05 \mathrm{eV}\times 100=5 \mathrm{eV}$

Now using

$\mathrm{E}=\varphi +\mathrm{K}.\mathrm{E}$

$5=3+\mathrm{K}.\mathrm{E}.$

$\mathrm{K}.\mathrm{E}.=2\mathrm{eV}=3.2\times {10}^{-19} \mathrm{J}$

for de Broglie wavelength $\lambda =\frac{\mathrm{h}}{\mathrm{mv}}$

$\mathrm{K}.\mathrm{E}.=\frac{1}{2}{\mathrm{mv}}^2$

so $\mathrm{v}=\sqrt{\frac{2\mathrm{K}.\mathrm{E}.}{\mathrm{m}}}$

hence $\mathrm{\lambda }=\frac{\mathrm{h}}{\sqrt{2\mathrm{K}.\mathrm{E}.\times \mathrm{m}}}$

$=\frac{6.63\times {10}^{-34}}{\sqrt{2\times 3.2\times {10}^{-19}\times 9.1\times {10}^{-31}}}$

$=\frac{6.63}{7.6}\times \frac{{10}^{-34}}{{10}^{-25}}=\frac{66.3\times {10}^{-10} \mathrm{m}}{7.6}$

$=8.72\times {10}^{-10} \mathrm{m}$

$\approx 9\times {10}^{-10} \mathrm{m}$

$=9 \stackrel{\mathrm{o}}{\mathrm{A}}$