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A 5 watt source emits monochromatic light of wavelength 5000 Å. 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
8
B
16
C
2
D
4
Detailed Solution
The number of photoelectrons emitted per second is proportional to the intensity of the incident light: $N_E \propto$ intensity.
For a point source, intensity $\propto \dfrac{1}{(\text{distance})^2}$
$\therefore N_E \propto \dfrac{1}{d^2}$
$\dfrac{N_2}{N_1} = \left(\dfrac{d_1}{d_2}\right)^2 = \left(\dfrac{0.5}{1.0}\right)^2 = \dfrac{1}{4}$
Therefore, as the distance is doubled, the number of photoelectrons decreases to $\dfrac{1}{4}$ of its value, i.e. it is reduced by a factor of 4.
The wavelength is unchanged, so the energy of each photoelectron is not affected.
For a point source, intensity $\propto \dfrac{1}{(\text{distance})^2}$
$\therefore N_E \propto \dfrac{1}{d^2}$
$\dfrac{N_2}{N_1} = \left(\dfrac{d_1}{d_2}\right)^2 = \left(\dfrac{0.5}{1.0}\right)^2 = \dfrac{1}{4}$
Therefore, as the distance is doubled, the number of photoelectrons decreases to $\dfrac{1}{4}$ of its value, i.e. it is reduced by a factor of 4.
The wavelength is unchanged, so the energy of each photoelectron is not affected.