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NCERT Class XII Chapter
Dual Nature of Radiation and Matter
Questions With Solutions

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Question : 2
Total: 37
The work function of caesium metal is 2.14 eV. When light of frequency 6 × 1014 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 W0 = 2.14 eV, υ = 6 × 1014 Hz
(a) Kmax = hυ – W0 = 6.63 × 10–34 × 6 × 1014 J – 2.14 eV
=
6.63×6×10−20
1.6×1019
 eV - 2.14 eV = 2.48 - 2.14 = 0.34 eV
(b) As eV0 = Kmax 0.34 eV
∴ Stopping potential, V0 = 0.34 V.
(c) Kmax =
1
2
mvmax2 = 0.34 eV = 0.34 × 1.6 × 10−19 J
or vmax2 =
2×0.34×1.6×10−19
m
 =
2×0.34×1.6×10−19
9.1×10−31
 = 119560.4 × 106
or vmax = 345.8 × 103ms−1 = 345.8 km s−1
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