Physics, asked by QueenUniversal1224, 10 months ago

Radiation of wavelength lambda is incident on a photocell.the fastest emitted electron has a speed v.if the wavelength is vhanged to 3lambda/4,the sleed of the fastest emitted electron will be A)>v(4/3)^1/2
B)

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Answered by utkarshgupta187
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Answered by wads
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Answer:

Radiation of wavelength λ is incident on a photocell. The fastest emitted electron has a speed v. If the wavelength is changed to 43λ, the speed of the fastest emitted electron will be : A .

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