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When a photosensitive surface is irradiated by lights of wavelengths λ1 and λ2, kinetic energies of emitted photoelectrons are E1 and E2 respectively. The work function of the photosensitive -

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प्रश्न

When a photosensitive surface is irradiated by lights of wavelengths `lambda_1` and `lambda_2`, kinetic energies of emitted photoelectrons are E1 and E2 respectively. The work function of the photosensitive surface is ____________.

विकल्प

  • `(lambda_2 "E"_1 + lambda_2 "E"_2)/(lambda_2 - lambda_1)`

  • `(lambda_2 "E"_2 + lambda_1 "E"_1)/(lambda_2 - lambda_1)`

  • `(lambda_1 "E"_1 + lambda_2 "E"_2)/(lambda_2 - lambda_1)`

  • `(lambda_1 "E"_1 + lambda_2 "E"_2)/(lambda_2 - lambda_1)`

MCQ
रिक्त स्थान भरें

उत्तर

When a photosensitive surface is irradiated by lights of wavelengths `lambda_1` and `lambda_2`, kinetic energies of emitted photoelectrons are E1 and E2 respectively. The work function of the photosensitive surface is `(lambda_1 "E"_1 + lambda_2 "E"_2)/(lambda_2 - lambda_1)`.

Explanation:

We know,

`"hv"_1 = "E"_1 + omega_0`

`therefore "hc"/lambda_1 = "E"_1 + omega_0`

`"hc" = ("E"_1 + omega_0) lambda_1`    ....(1)

`"Similarly, hc" = ("E"_2 + omega_0)lambda_2`    ....(2)

`therefore omega_0 lambda_1 + "E"_1 lambda_1 = omega_0 lambda_2 + "E"_2 lambda_2`

`omega_0 (lambda_1 - lambda_2) = "E"_2 lambda_2 - "E"_1 lambda_1`

`omega_0 = ("E"_2 lambda_2 - "E"_1 lambda_1)/(lambda_1 - lambda_2)`

`= ("E"_1 lambda_1 - "E"_2 lambda_2)/(lambda_2 - lambda_1)`

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The Photoelectric Effect
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