CBSE Class 12 · 2024 · Modern PhysicsShort Answer

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A photosensitive surface of work function 2.1 eV is irradiated by radiation of wavelength 150 nm. Calculate
(i) the threshold wavelength,
(ii) the energy (in eV) of an incident photon, and
(iii) the maximum kinetic energy of the emitted photoelectron.

Answer: $λ_{0} \approx 590$ nm; $E \approx 8.27$ eV; $K_{ma x} \approx 6.17$ eV.

Solution

(i) Threshold wavelength:

λ_{0} = \frac{1240 eV nm}{ϕ _{0}} = \frac{1240}{2.1} \approx 590 nm .

(ii) Energy of the incident photon:

E = \frac{1240}{150} \approx 8.27 eV .

(iii) Maximum kinetic energy:

K_{ma x} = E - ϕ_{0} = 8.27 - 2.1 \approx 6.17 eV .

λ_{0} \approx 590 nm, E \approx 8.27 eV, K_{ma x} \approx 6.17 eV

CBSE Class 12 Board · 2024 · Subjective

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