Question

A $ 5 \,W $ source emits monochromatic light of wavelength $ 5000 \,\mathring{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

• A. $ 4 $
• B. $ 8 $
• C. $ 16 $
• D. $ 2 $

Answer 1

Answer: (A)

Intensity of light is inversely proportional to square of distance.
ie, $I \propto \frac{1}{r^{2}}$
or $\frac{I_{2}}{I_{1}}=\frac{\left(r_{1}\right)^{2}}{\left(r_{2}\right)^{2}}$
Given, $r_{1}=0.5\,m$, $r_{2}=1.0\,m$
Therefore, $\frac{I_{2}}{I_{1}}=\frac{\left(0.5\right)^{2}}{\left(1\right)^{2}}$
$=\frac{1}{4}$
Now, since number of photoelectrons emitted per second is directly proportional to intensity, so number of electrons emitted would decrease by factor of $4$.