A 5 watt source emits monochromatic light of wavelength 5000 A when placed 0.5 m away, it liberates photoelectric from a photosensitive metallic surface. When the source is moved to a distance of 1.0 m, The numbers of photo electrons liberated will:

be reduced by a factor of 2

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be reduced by a factor of 4

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be reduced by a factor of 8

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be reduced by a factor of 16

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Solution

The correct option is B be reduced by a factor of 4

$\begin{matrix} 1 \propto \frac{1}{r^{2}} \frac{l_{1}}{l_{2}} = \frac{{(r_{1})}^{2}}{{(r_{2})}^{2}} G i v e n, r_{1} = 0.5 m, r_{2} = 1.0 m t h e r e f o r e, \frac{l_{2}}{l_{1}} = \frac{{(0.5)}^{2}}{{(1)}^{2}} = \frac{1}{4} \end{matrix}$
Now, since number of photoelectric emitted per second is directly
pr optional intensity, so number of electron emitted would decreased by factor of $4$ .

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A 5 watt source emits monochromatic light of wavelength 5000 A when placed 0.5 m away, it liberates photoelectric from a photosensitive metallic surface. When the source is moved to a distance of 1.0 m, The numbers of photo electrons liberated will:

The correct option is B be reduced by a factor of 41∝1r2l1l2=(r1)2(r2)2Given,r1=0.5m,r2=1.0mtherefore,l2l1=(0.5)2(1)2=14Now, since number of photoelectric emitted per second is directly pr optional intensity, so number of electron emitted would decreased by factor of 4.