First slide

Quantum Theory of Light

Question

A 5 W source emits monochromatic light of wavelength 5000 A. 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

Moderate

A

4
B

8
C

16
D

2

Solution

Intensity of light is inversely proportional to square of distance.
$\begin{array}{l} That is, I \propto \frac{1}{r^{2}} or \frac{I_{2}}{I_{1}} = \frac{{(r_{1})}^{2}}{{(r_{2})}^{2}} \\ Given, r_{1} = 0 .5 m, r_{2} = 1 .0 m \\ Therefore \frac{I_{2}}{I_{1}} = \frac{{(0 .5)}^{2}}{{(1)}^{2}} = \frac{1}{4} \end{array}$
Now, since number of photoelectrons emitted per second is directly proportional to intensity, so number of electrons emitted would decrease by factor of 4.

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