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प्रश्न
Monochromatic light of wavelength 396 nm is incident on the surface of a metal whose work function is 1.125 eV. Calculate:
- the energy of an incident photon in eV.
- the maximum kinetic energy of photoelectrons in eV.
उत्तर
(i) Energy of proton = `"hc"/lambda`
`= (6.6 xx 10^-34 xx 3 xx 10^8)/(396 xx 10^-9)`
= 5 × 10-19 J
= 3.125 eV
(ii) `"E"_"max" = "hc"/lambda - "W"`
= 3.125 - 1.125
= 2 eV.
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