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NCERT Class XII Chapter
Dual Nature of Radiation and Matter
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Question : 2 of 37
Marks: +1, -0
The work function of caesium metal is 2.14 eV. When light of frequency 6 × 101410^{14} Hz is incident on the metal surface, photoemission of electrons occurs. What is the :
(a) maximum kinetic energy of the emitted electrons,
(b) stopping potential, and
(c) maximum speed of the emitted photo-electrons?
Solution:  
Here W0W_0 = 2.14 eV, υ = 6 × 1014 Hz
(a) Kmax⁡K_{\max} = hυ – W0W_0 = 6.63 × 10−3410^{-34} × 6 × 101410^{14} J – 2.14 eV
= 6.63×6×10−201.6×1019\frac{6.63 \times 6 \times 10^{-20}}{1.6 \times 10^{19}} eV - 2.14 eV = 2.48 - 2.14 = 0.34 eV
(b) As eV0\mathrm{eV}_0 = Kmax⁡K_{\max} 0.34 eV
∴ Stopping potential, V0V_0 = 0.34 V.
(c) Kmax⁡K_{\max} = 12mvmax⁡2\frac{1}{2} m v^{2}_{\max} = 0.34 eV = 0.34 × 1.6 × 10−1910^{-19} J
or vmax⁡2v_{\max}^{2} = 2×0.34×1.6×10−19m\frac{2 \times 0.34 \times 1.6 \times 10^{-19}}{m} = 2×0.34×1.6×10−199.1×10−31\frac{2 \times 0.34 \times 1.6 \times 10^{-19}}{9.1 \times 10^{-31}} = 119560.4 × 10610^{6}
or vmax⁡v_{\max} = 345.8 × 103 ms−110^{3}\,\mathrm{ms}^{-1} = 345.8 km s−1\mathrm{s}^{-1}
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