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A 5 W source emits monochromatic light of wavelength 5000 ångström. When placed 0.5 m away, it liberates photoelectrons from a photosensitive metallic surface. When the source is moved to a distance of 1.0 m, the number of photoelectrons liberated will be reduced by a factor of:

Asked in AIPMT 2007 · Intensity, area and distance

Answer: (4) 4

Step-by-step solution

The number of photoelectrons per second follows the intensity of the light at the surface.

For a point source, I=P/(4π d²), so I∝1/(d²).

Doubling the distance from 0.5 m to 1.0 m gives (I₁)/(I₂)=((1.0)/(0.5))²=4.

So the intensity, and with it the number of photoelectrons, falls by a factor of 4.

Neither the 5 W nor the wavelength is needed: the factor depends only on the distance ratio.

Why the other options are wrong

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