The work function of a certain metal is `(hC)/lambda_(0)`. When a monochromatic light of wavelength `lambda lt lambda_(0)` is incident such that the plate gains a total power `P`. If the efficiency of photoelectric emission is `eta%` and all the emitted photoelectrons are captured by a hollow conducting sphere of radius `R` already charged to potential `V`, then neglecting any interaction of potential of the sphere at time `t` is:

Options

A. `V+(100etalambdaPet)/(4piepsilonRhC)`
B. `V-(etalambdaPet)/(400piepsilonRhC)`
C. V
D. `(lambdaPet)/(4piepsilonRhC)`

Answer

To solve the problem, we need to follow a series of steps to derive the potential of the hollow conducting sphere after capturing the emitted photoelectrons.

### Step-by-Step Solution:

1. **Identify the Energy of a Photon:**
The energy of a single photon is given by the equation:
\[ E = \frac{hc}{\lambda} \]
where \( h \) is Planck's constant, \( c \) is the speed of light, and \( \lambda \) is the wavelength of the incident light.

2. **Calculate the Number of Photons Incident per Second:**
The power \( P \) of the light source is the energy per unit time. Therefore, the number of photons \( N \) incident per second can be calculated as:
\[ N = \frac{P}{E} = \frac{P}{\frac{hc}{\lambda}} = \frac{P \lambda}{hc} \]

3. **Determine the Number of Emitted Photoelectrons:**
Given that the efficiency of photoelectric emission is \( \eta\% \), the number of emitted photoelectrons per second is:
\[ n = \frac{\eta}{100} \times N = \frac{\eta}{100} \times \frac{P \lambda}{hc} \]

4. **Calculate the Charge Captured by the Sphere:**
The charge \( Q \) of the emitted photoelectrons captured by the sphere in time \( t \) is:
\[ Q = n \times e \times t = \left(\frac{\eta}{100} \times \frac{P \lambda}{hc}\right) \times e \times t \]
where \( e \) is the charge of an electron.

5. **Determine the New Potential of the Sphere:**
The potential \( V' \) of the sphere after capturing the charge is given by:
\[ V' = V - \frac{Q}{4 \pi \epsilon_0 R} \]
Substituting the expression for \( Q \):
\[ V' = V - \frac{\left(\frac{\eta}{100} \times \frac{P \lambda}{hc} \times e \times t\right)}{4 \pi \epsilon_0 R} \]

6. **Final Expression for the New Potential:**
Simplifying the expression gives:
\[ V' = V - \frac{\eta \lambda P e t}{400 \pi \epsilon_0 R hc} \]

### Final Answer:
The potential of the sphere at time \( t \) is:
\[ V' = V - \frac{\eta \lambda P e t}{400 \pi \epsilon_0 R hc} \]