When light of wavelength 400nm is incident on the cathode of photocell, the stopping potential recorded is 6V. If the wavelength of the incident light is to 600nm, calculate the new stopping potential. [Given `h=6.6xx10^(-34) Js, c=3xx10^(8)m//s , e=1.6xx10^(-19)C`]

Options

A. `1.03 V`
B. `2.42 V`
C. `4.97 V`
D. `3.58 V`

Answer

To solve the problem, we will use Einstein's photoelectric equation, which relates the energy of the incident photons to the stopping potential in a photocell. The equation is given as:

\[ eV = \frac{hc}{\lambda} - W_0 \]

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 light,
- \( W_0 \) is the work function of the material.

### Step 1: Write down the given values
From the problem:
- For the first wavelength (\( \lambda_1 = 400 \, \text{nm} = 400 \times 10^{-9} \, \text{m} \)):
- Stopping potential (\( V_1 = 6 \, \text{V} \))

- For the second wavelength (\( \lambda_2 = 600 \, \text{nm} = 600 \times 10^{-9} \, \text{m} \)):
- We need to find \( V_2 \).

### Step 2: Write the equations for both wavelengths
Using the photoelectric equation for both wavelengths, we have:

1. For \( \lambda_1 \):
\[ eV_1 = \frac{hc}{\lambda_1} - W_0 \]
\[ e(6) = \frac{(6.6 \times 10^{-34})(3 \times 10^8)}{400 \times 10^{-9}} - W_0 \]

2. For \( \lambda_2 \):
\[ eV_2 = \frac{hc}{\lambda_2} - W_0 \]
\[ eV_2 = \frac{(6.6 \times 10^{-34})(3 \times 10^8)}{600 \times 10^{-9}} - W_0 \]

### Step 3: Subtract the two equations
By subtracting the two equations, we eliminate \( W_0 \):

\[ eV_1 - eV_2 = \left(\frac{hc}{\lambda_1} - W_0\right) - \left(\frac{hc}{\lambda_2} - W_0\right) \]

This simplifies to:

\[ e(V_1 - V_2) = \frac{hc}{\lambda_1} - \frac{hc}{\lambda_2} \]

### Step 4: Rearranging the equation
Rearranging gives us:

\[ V_2 = V_1 - \frac{hc}{e} \left(\frac{1}{\lambda_1} - \frac{1}{\lambda_2}\right) \]

### Step 5: Substitute the known values
Now, substitute the known values into the equation:

- \( h = 6.6 \times 10^{-34} \, \text{Js} \)
- \( c = 3 \times 10^8 \, \text{m/s} \)
- \( e = 1.6 \times 10^{-19} \, \text{C} \)
- \( \lambda_1 = 400 \times 10^{-9} \, \text{m} \)
- \( \lambda_2 = 600 \times 10^{-9} \, \text{m} \)
- \( V_1 = 6 \, \text{V} \)

Calculating \( \frac{hc}{e} \):

\[ \frac{hc}{e} = \frac{(6.6 \times 10^{-34})(3 \times 10^8)}{1.6 \times 10^{-19}} \approx 1.237 \times 10^{-6} \, \text{V m} \]

Now, substituting into the equation for \( V_2 \):

\[ V_2 = 6 - \left(1.237 \times 10^{-6}\right) \left(\frac{1}{400 \times 10^{-9}} - \frac{1}{600 \times 10^{-9}}\right) \]

Calculating the term in parentheses:

\[ \frac{1}{400 \times 10^{-9}} - \frac{1}{600 \times 10^{-9}} = \frac{600 - 400}{240000} = \frac{200}{240000} = \frac{1}{1200} \]

Thus:

\[ V_2 = 6 - \left(1.237 \times 10^{-6} \times \frac{1}{1200}\right) \]

Calculating \( 1.237 \times 10^{-6} \times \frac{1}{1200} \):

\[ \approx 1.031 \times 10^{-9} \, \text{V} \]

Finally, calculating \( V_2 \):

\[ V_2 \approx 6 - 0.00497 \approx 4.97 \, \text{V} \]

### Final Answer:
The new stopping potential \( V_2 \) is approximately **4.97 V**.

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