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. a.`V+(100eta lambda Pet)/(4piepsilon_(0)RhC)`
B. b.`V-(etalambdaPet)/(400pi epsilon_(0)RhC)`
C. c..`V`
D. d.`(lambdaPet)/(4 pi epsilon_(0)RhC)`

Answer

To solve the problem step by step, we need to analyze the given information and apply the relevant physics concepts.

### Step 1: Understand the Work Function
The work function \( \phi \) of the metal is given by:
\[ \phi = \frac{hc}{\lambda_0} \]
where \( h \) is Planck's constant, \( c \) is the speed of light, and \( \lambda_0 \) is the threshold wavelength.

### Step 2: Calculate the Energy of Incident Photons
For monochromatic light of wavelength \( \lambda \) (where \( \lambda < \lambda_0 \)), the energy \( E \) of a single photon is given by:
\[ E = \frac{hc}{\lambda} \]

### Step 3: Determine the Number of Photons Incident Per Second
The power \( P \) of the light incident on the plate can be expressed in terms of the energy of the photons and the number of photons \( n \) emitted per second:
\[ P = n \cdot E \]
Substituting the expression for \( E \):
\[ P = n \cdot \frac{hc}{\lambda} \]
From this, we can solve for \( n \):
\[ n = \frac{P \lambda}{hc} \]

### Step 4: Calculate the Number of Photoelectrons Emitted
Given that the efficiency of photoelectric emission is \( \eta\% \), the number of photoelectrons emitted per second \( n_e \) is:
\[ n_e = \frac{\eta}{100} \cdot n = \frac{\eta}{100} \cdot \frac{P \lambda}{hc} \]

### Step 5: Determine the Charge Collected by the Sphere
The total charge \( Q \) collected by the hollow conducting sphere after time \( t \) is given by:
\[ Q = n_e \cdot e \cdot t \]
Substituting the expression for \( n_e \):
\[ Q = \left( \frac{\eta}{100} \cdot \frac{P \lambda}{hc} \right) \cdot e \cdot t \]

### Step 6: Relate Charge to Potential Change
The potential \( V \) of a charged sphere is given by:
\[ V' = V - \frac{Q}{4 \pi \epsilon_0 R} \]
Substituting for \( Q \):
\[ V' = V - \frac{1}{4 \pi \epsilon_0 R} \left( \frac{\eta}{100} \cdot \frac{P \lambda}{hc} \cdot e \cdot t \right) \]

### Step 7: Final Expression for Potential
Thus, the potential \( V' \) after time \( t \) is:
\[ V' = V - \frac{\eta P \lambda e t}{400 \pi \epsilon_0 hc R} \]

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