The work function of caesium metal is 2.14 eV. When light of frequency 6 x 10¹⁴Hz is incident on the metal surface, photoemission of electrons occurs. What is the maximum kinetic energy of the emitted electrons?

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Solution

Given,
Work function of caesium metal = 2.14 eV
Frequency of light,

ν

= 6 x 10¹⁴Hz

Maximum kinetic energy of the emitted electrons,

K . E ._{\max} = hv - ϕ_{0} = \frac{6.63 \times 10^{- 34} \times 6 \times 10^{14}}{1.6 \times 10^{- 19}} - 2.14

E_{\max} = 0.346 eV = 0.35 eV

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