Question

Light of frequency f and intensity I falls on cathode of tube having work function $\varphi$ due to which photocurrent is generated in the tube. List - 1 gives some quantities and List - 2 gives changes in these quantities when some modification is done on incident light.



$\begin{array}{cccc}& \text{List-1}& & \text{List-2}\\ \text{I}& \text{Stopping potential}& \text{P}& \text{Increases}\\ \text{II}& \text{Saturated current}& \text{Q}& \text{Decreases}\\ \text{III}& \text{Minimum de Broglie wavelength of photoelectron}& \text{R}& \text{Remains same}\\ \text{IV}& \text{Work function of cathode}& \text{S}& \text{Becomes more than double}\\ & & \text{T}& \text{Becomes half}\\ & & \text{U}& \text{None}\end{array}$

If intensity of light is doubled keeping frequency constant, match enteries in List - 1 with those in List - 2.

• A. I $\to$ P II $\to$ P III $\to$ Q, T IV $\to$ R
• B. I $\to$ P II $\to$ R III $\to$ P IV $\to$ R
• C. I $\to$ R II $\to$ P III $\to$ R IV $\to$ R
• D. I $\to$ R II $\to$ P,S III $\to$ Q IV $\to$ R

Answer 1

The correct option is C I $\to$ R II $\to$ P III $\to$ R IV $\to$ R

Intensity is doubled but frequency is constant. This means, the photon flux is doubled. So, saturation current gets doubled. The work function has no relation with the photon, so it remains the same. The stopping potential also remains same as the energy of each photon remains same. The threshold wavelength also remains the same as the work function is same.