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Question
- Calculate the frequency of a photon of energy 6.5 × 10−19 J.
- Can this photon cause the emission of an electron from the surface of Cs of work function 2.14 eV? If yes, what will be the maximum kinetic energy of the photoelectron?
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Solution
(a) Frequency of photon = `"E"/"h"`
∴ Frequency = v = `(6.5 xx 10^-19)/(6.6 xx 10^-34)` = 1015 Hz
(b) Work function of Cs = Φ0 = 2.14 eV
Threshold frequency = v0 = `Φ_0/"h"`
= `(2.14 xx 1.6 xx 10^-19)/(6.6 xx 10^-34)`
= 0.5 × 1015 Hz
Since the energy of an incident photon is more than the threshold frequency, the emission of photoelectrons will be possible.
KEmax = Energy of incident photon - Work function
KEmax = 6.5 × 10−19 − 2.14 × 1.6 × 10−19
∴ KEmax = 3.1 × 10−19 J
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