A photo cell is receiving light from a source placed at a distance of 1 m- If the same source is to be placed at a distance of 2 m- then the ejected electronA- Moves with one-fourth energy as that of the initial energyB- Moves with one-fourth of momentum as that of the initial momentumC- Will be half in numberD- Will be one-fourth in number

The correct option is D Will be one fourth in numberNumber of ejected electrons ∞ (Intensity) ∞ 1/(Distance)^2 Therefore an increment of distance two times wil ...

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Standard XII

Physics

Intensity in Photon Model

A photo cell ...

Question

A photo cell is receiving light from a source placed at a distance of 1 m. If the same source is to be placed at a distance of 2 m, then the ejected electron

Moves with one-fourth energy as that of the initial energy

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Moves with one-fourth of momentum as that of the initial momentum

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Will be half in number

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Will be one-fourth in number

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Solution

The correct option is D Will be one-fourth in number
Number of ejected electrons ∞ (Intensity) ∞ $\frac{1}{(D i s t a n c e)^{2}}$
Therefore an increment of distance two times will reduce the number of ejected electrons to $\frac{1}{4} t h$ of the previous one.

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A photo cell is receiving light from a source placed at a distance of 1 m. If the same source is to be placed at a distance of 2 m, then the ejected electron

The correct option is D Will be one-fourth in numberNumber of ejected electrons ∞ (Intensity) ∞ 1(Distance)2 Therefore an increment of distance two times will reduce the number of ejected electrons to 14th of the previous one.

The correct option is D Will be one-fourth in number
Number of ejected electrons ∞ (Intensity) ∞ $\frac{1}{(D i s t a n c e)^{2}}$
Therefore an increment of distance two times will reduce the number of ejected electrons to $\frac{1}{4} t h$ of the previous one.