When photon of wavelength `lambda_(1)` are incident on an isolated shere supended by an insulated , the corresponding stopping potential is found to be `V`. When photon of wavelength `lambda_(2)` are used , the orresponding stopping potential was thrice the above value. If light of wavelength `lambda_(3)` is used , carculate the stopping potential for this case.

Options

A. `(hc)/(e)((1)/(lambda_(3))+(1)/(2lambda_(2))-(1)/(lambda_(1)))`
B. `(hc)/(e)((1)/(lambda_(3))+(1)/(2lambda_(2))-(1)/(2lambda_(1)))`
C. `(hc)/(e)((1)/(lambda_(3))+(1)/(lambda_(2))+(1)/(lambda_(1)))`
D. `(hc)/(e)((1)/(lambda_(3))-(1)/(lambda_(2))-(1)/(lambda_(1)))`

Answer

To solve the problem, we need to use the photoelectric effect equation, which relates the stopping potential (V), the wavelength of the incident photon (λ), the work function (φ), and Planck's constant (h) along with the speed of light (c). The equation is given by:

\[ eV = \frac{hc}{\lambda} - \phi \]

Where:
- \( e \) is the charge of an electron,
- \( V \) is the stopping potential,
- \( h \) is Planck's constant,
- \( c \) is the speed of light,
- \( \lambda \) is the wavelength of the incident photon,
- \( \phi \) is the work function of the material.

### Step 1: Write the equations for the two given wavelengths
For the first wavelength \( \lambda_1 \):
\[ eV = \frac{hc}{\lambda_1} - \phi \] (Equation 1)

For the second wavelength \( \lambda_2 \), where the stopping potential is three times the first:
\[ e(3V) = \frac{hc}{\lambda_2} - \phi \] (Equation 2)

### Step 2: Rearranging the equations
From Equation 1:
\[ \phi = \frac{hc}{\lambda_1} - eV \]

Substituting this expression for \( \phi \) into Equation 2 gives:
\[ 3eV = \frac{hc}{\lambda_2} - \left(\frac{hc}{\lambda_1} - eV\right) \]

### Step 3: Simplifying the equation
Now, rearranging the equation:
\[ 3eV + eV = \frac{hc}{\lambda_2} - \frac{hc}{\lambda_1} \]
\[ 4eV = \frac{hc}{\lambda_2} - \frac{hc}{\lambda_1} \]

### Step 4: Solve for \( V \)
Factoring out \( hc \):
\[ 4eV = hc \left(\frac{1}{\lambda_2} - \frac{1}{\lambda_1}\right) \]
\[ V = \frac{hc}{4e} \left(\frac{1}{\lambda_2} - \frac{1}{\lambda_1}\right) \]

### Step 5: Finding the work function
Now, we can express the work function \( \phi \) using the value of \( V \):
Substituting \( V \) back into the equation for \( \phi \):
\[ \phi = \frac{hc}{\lambda_1} - eV \]
\[ \phi = \frac{hc}{\lambda_1} - e \left(\frac{hc}{4e} \left(\frac{1}{\lambda_2} - \frac{1}{\lambda_1}\right)\right) \]
\[ \phi = \frac{hc}{\lambda_1} - \frac{hc}{4} \left(\frac{1}{\lambda_2} - \frac{1}{\lambda_1}\right) \]

### Step 6: Calculate the stopping potential for \( \lambda_3 \)
Now, we need to find the stopping potential \( V' \) for the wavelength \( \lambda_3 \):
\[ eV' = \frac{hc}{\lambda_3} - \phi \]

Substituting the expression for \( \phi \):
\[ eV' = \frac{hc}{\lambda_3} - \left(\frac{hc}{\lambda_1} - \frac{hc}{4} \left(\frac{1}{\lambda_2} - \frac{1}{\lambda_1}\right)\right) \]

### Step 7: Final expression for \( V' \)
This gives us the final expression for the stopping potential \( V' \):
\[ V' = \frac{hc}{e} \left(\frac{1}{\lambda_3} - \left(\frac{1}{\lambda_1} - \frac{1}{4} \left(\frac{1}{\lambda_2} - \frac{1}{\lambda_1}\right)\right)\right) \]

### Conclusion
Thus, the stopping potential for the wavelength \( \lambda_3 \) can be calculated using the above expression.