Chemistry - Atomic Structure Question with Solution | TestHub

Total energy $\left(\mathrm{E}\right)$ of any photon is given by the relation : $\mathrm{E}=\frac{\mathrm{hc}}{\mathrm{\lambda }}$

Where, $\mathrm{h}=$ Planck's constant $=6.6\times {10}^{-34} \mathrm{Js}$

$\mathrm{c}=$ Velocity of light $=3\times {10}^8 \mathrm{m}/\mathrm{s}$

$\mathrm{\lambda }=$ wavelength $=1.5\times {10}^{-10} \mathrm{m},$ i.e, $\left(150 \mathrm{pm}\right)$

Thus, $\mathrm{E}=\frac{\mathrm{hc}}{\mathrm{\lambda }}=\frac{6.6\times {10}^{-34}\times 3\times {10}^8}{1.5\times {10}^{-10}}; \mathrm{E}=1.32\times {10}^{-15} \mathrm{J}$ and, energy of ejected electron $\left({\mathrm{E}}^{\text{'}}\right)$

${\mathrm{E}}^{\text{'}}=\frac{1}{2}{\mathrm{mv}}^2$,

where,

$\mathrm{m}=$ mass of electron $=9.1\times {10}^{-31} \mathrm{kg}$

$\mathrm{v}=$ velocity of electron

$=1.5\times {10}^7 \mathrm{m}/\mathrm{s}$

${\mathrm{E}}^{\text{'}}=\frac{1}{2}{\mathrm{mv}}^2=\frac{1}{2}\times 9.1\times {10}^{-31}\times {\left(1.5\times {10}^7\right)}^2$

${\mathrm{E}}^{\text{'}}=1.024\times {10}^{-16}$

Thus, total energy of photon $=$ Binding energy of electron

$\left(\mathrm{B}\right)$ + energy of ejected electron $\left({\mathrm{E}}^{\text{'}}\right)$

Thus, $1.32\times {10}^{-15}=\left(\mathrm{B}\right)+\left(\mathrm{E}\right)$

$\therefore \left(\mathrm{B}\right)=\left(\mathrm{E}\right)-\left({\mathrm{E}}^{\text{'}}\right)=\left[1.32\times {10}^{-15}\right]-\left[1.024\times {10}^{-16}\right]$

$=1.2176\times {10}^{-15} \mathrm{J}$

$=\frac{1.2176\times {10}^{-15}}{1.6\times {10}^{-19}} \mathrm{eV}=7.6\times {10}^3 \mathrm{eV}$