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(A) The intensity of light $I$ at a distance $r$ from a point source is given by $I = \frac{P_0}{4 \pi r^2}$, where $P_0$ is the power of the source.S

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be reduced by a factor of $4$

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(A) The intensity of light $I$ at a distance $r$ from a point source is given by $I = \frac{P_0}{4 \pi r^2}$, where $P_0$ is the power of the source.Since the number of photoelectrons liberated per second is directly proportional to the number of photons incident on the surface, and the number of photons is proportional to the intensity of light, we have $N \propto I$.Therefore, $N \propto \frac{1}{r^2}$.Given $r_1 = 0.5\; m$ and $r_2 = 1.0\; m$, the ratio of the number of photoelectrons is:$\frac{N_2}{N_1} = \frac{r_1^2}{r_2^2} = \left( \frac{0.5}{1.0} \right)^2 = \left( \frac{1}{2} \right)^2 = \frac{1}{4}$.Thus, the number of photoelectrons liberated is reduced by a factor of $4$.

$A$ $5\; W$ 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

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