If the photon of the wavelength 150 pm strikes an atom and one of its inner bond electrons is
ejected out with a velocity of ${1.5 × 10}^{7} {ms}^{-1}$ , what is the energy with which it is bond to the
nucleus?

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${7.6 × 10}^{3} eV$

${8.12 × 10}^{3} eV$

${2.15 × 10}^{3} eV$

$12 \times 10^{2} eV$

answer is C.

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Detailed Solution

Energy of photon, $E = \frac{hc}{λ} = \frac{{66 × 10}^{-34} {× 3 × 10}^{8}}{{1.5 × 10}^{-10}}$

${= 1.32 × 10}^{15} J$

Energy of ejected electron, $E = \frac{1}{2} {mv}^{2} = \frac{1}{2} {× 9.1 × 10}^{-31} × {({1.5 × 10}^{7})}^{2}$

${= 1.024 × 10}^{-16}$

Total energy of photon = binding energy of electron + energy of ejected electron ${1.32 × 10}^{-15}$ ${= binding energy + 1.024 × 10}^{-16}$

$∴ Binding energy$

$= ({1.32 × 10}^{-15}) - ({1.024 × 10}^{-16})$

${= 1.2176 × 10}^{-15} J = \frac{{1.2176 × 10}^{-15}}{{1.6 × 10}^{-19}} eV$

${= 7.6 × 10}^{3} eV$

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