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NCERT Class XI Chemistry Structure of Atom Solutions

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Question : 9 of 67
Marks: +1, -0
A photon of wavelength 4 × 10−710^{-7} m strikes on metal surface, the work function of the metal being 2.13 eV. Calculate (i) the energy of the photon (eV), (ii) the kinetic energy of the emission, and (iii) the velocity of the photoelectron (1 eV = 1.6020 × 10−1910^{-19} J).
Solution:  
(i) Energy of photon (E) = hcλ\frac{hc}{\lambda} = 6.626×10−34×3×1084×10−7\frac{6.626 \times 10^{-34} \times 3 \times 10^8}{4 \times 10^{-7}} = 4.9695 × 10−19J10^{-19} \text{J}
E = 4.9695×10−191.6020×10−19\frac{4.9695 \times 10^{-19}}{1.6020 \times 10^{-19}} = 3.102 eV
(ii) Kinetic energy of the emission = energy of photon – work function
= (3.102 – 2.13) eV = 0.972 eV
1 eV = 1.6020 × 10−19J10^{-19} \text{J}
0.972 eV = 1.6020 × 10−1910^{-19} × 0.972 J = 1.557 × 10−19J10^{-19} \text{J}
(iii) K.E. = 12mv2\frac{1}{2} m v^2
Velocity of the photoelectron (v) = 2K.E.m\sqrt{\frac{2 \text{K.E.}}{m}}
v = 2×1.557×10−199.1×10−31\sqrt{\frac{2 \times 1.557 \times 10^{-19}}{9.1 \times 10^{-31}}} = 5.85 × 105m/s10^5 \text{m/s}
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