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प्रश्न
When a 6000 Å light falls on the cathode of a photo cell, photoemission takes place. If a potential of 0.8 V is required to stop emission of electron, then determine the
- frequency of the light
- energy of the incident photon
- work function of the cathode material
- threshold frequency and
- net energy of the electron after it leaves the surface.
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उत्तर
Given: λ = 6000 × 10−10 m
eV0 = 0.8 eV
i. v = `"C"/λ`
= `(3 xx 10^8)/(6 xx 10^-7)`
= 0.5 × 1015
v = 5 × 1014 Hz
ii. E = hv
= `(6.626 xx 10^-34 xx 5 xx 10^14)/(1.6 xx 10^-19)`
= `53.13/1.6 xx 10^-20 xx 10^19`
E = 2.07 eV
iii. ev0 = `1/2` mv2 = 0.8 eV
W = hv – `1/2` mv2
= (2.07 – 0.8) eV
W = 1.27 eV
iv. W = hv0
v0 = `(1.27 xx 1.6 xx 10^-19)/(6.626 xx 10^-34)`
= `2.032/6.626 xx 10^15`
v0 = 3.06 × 1014 Hz
v. Net Energy = E = hv – hv0
E = (2.07 – 1.27) eV
E = 0.8 eV
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