CBSE Class 12 Physics 2018 Delhi Set 1 Paper

Question : 9

Total: 26

If light of wavelength 412.5‌nm is incident on each of the metals given below, which ones will show photoelectric emission and why?

Metal	Work Function (eV)
Na	1.92
K	2.15
Ca	3.20
Mo	4.17

Solution:

Calculating the energy of the incident photon 1 Identifying the metals
Reason
The energy of a photon of incident radiation is given by
‌E=‌

∴ ‌E=‌

6.63×10−34×3×108

(412.5×10−9)×(1.6×10−19)

‌eV
‌=3.01‌eV
Hence, only Na and K will show photoelectric emission
Reason: The energy of the incident photon is more than the work function of only these two metals.

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