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NCERT Class XI Chemistry Structure of Atom Solutions

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Question : 54 of 67
Marks: +1, -0
If the photon of the wavelength 150 pm strikes an atom and one of its inner bound electrons is ejected out with a velocity of 1.5 × 107 m s−110^{7}\,\text{m s}^{-1}, calculate the energy with which it is bound to the nucleus.
Solution:  
Photon of wavelength = 150 pm = 150 × 10−12 m10^{-12}\,\text{m}
Energy of photon (E) = hcλ\frac{hc}{\lambda} = 6.625×10−34×3×108150×1012\frac{6.625 \times 10^{-34} \times 3 \times 10^{8}}{150 \times 10^{12}} = 0.1325 × 10−1410^{-14}
= 13.25 × 10−16 J10^{-16}\,\text{J}
Energy of the ejected electron = 12mv2\frac{1}{2} m v^{2} = 12\frac{1}{2} × 9.11 × 10−31×(1.5×107)210^{-31} \times (1.5 \times 10^{7})^{2}
Energy with which the electron is bound to the nucleus
= (13.25 × 10−1610^{-16} – 1.025 × 10−1610^{-16}) J = 12.225 × 10−16 J10^{-16}\,\text{J}
= 12.225×10−161.602×10−19\frac{12.225 \times 10^{-16}}{1.602 \times 10^{-19}} = 7.63 × 10310^{3} eV
[Since 1.602 × 10−19 J10^{-19}\,\text{J} = 1 eV]
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