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`]

Explanation

To solve the problem, we will use Einstein's photoelectric equation, which relates the stopping potential (V₀), the wavelength of light (λ), the Planck's constant (h), the speed of light (c), and the work function (φ₀) of the material. The equation is given by:

\[ e \cdot V₀ = \frac{h \cdot c}{\lambda} - \phi₀ \]

Where:
- \( e \) is the charge of the electron,
- \( V₀ \) is the stopping potential,
- \( h \) is Planck's constant,
- \( c \) is the speed of light,
- \( \lambda \) is the wavelength of incident light,
- \( φ₀ \) is the work function of the material.

### Step 1: Write down the equations for the two wavelengths

For the first wavelength (λ₁ = 400 nm):
\[ e \cdot V_{0.1} = \frac{h \cdot c}{\lambda_1} - \phi₀ \]
Substituting the values:
\[ e \cdot 6 = \frac{(6.6 \times 10^{-34}) \cdot (3 \times 10^{8})}{400 \times 10^{-9}} - \phi₀ \]
This is our Equation 1.

For the second wavelength (λ₂ = 600 nm):
\[ e \cdot V_{0.2} = \frac{h \cdot c}{\lambda_2} - \phi₀ \]
Substituting the values:
\[ e \cdot V_{0.2} = \frac{(6.6 \times 10^{-34}) \cdot (3 \times 10^{8})}{600 \times 10^{-9}} - \phi₀ \]
This is our Equation 2.

### Step 2: Subtract Equation 2 from Equation 1

When we subtract Equation 2 from Equation 1, we eliminate φ₀:
\[ e \cdot V_{0.1} - e \cdot V_{0.2} = \left( \frac{h \cdot c}{\lambda_1} - \phi₀ \right) - \left( \frac{h \cdot c}{\lambda_2} - \phi₀ \right) \]

This simplifies to:
\[ e \cdot (V_{0.1} - V_{0.2}) = \frac{h \cdot c}{\lambda_1} - \frac{h \cdot c}{\lambda_2} \]

### Step 3: Rearranging the equation

Rearranging gives:
\[ V_{0.1} - V_{0.2} = \frac{h \cdot c}{e} \left( \frac{1}{\lambda_1} - \frac{1}{\lambda_2} \right) \]

### Step 4: Substitute the known values

Now, substituting the known values:
- \( V_{0.1} = 6 \, V \)
- \( h = 6.6 \times 10^{-34} \, Js \)
- \( c = 3 \times 10^{8} \, m/s \)
- \( e = 1.6 \times 10^{-19} \, C \)
- \( \lambda_1 = 400 \times 10^{-9} \, m \)
- \( \lambda_2 = 600 \times 10^{-9} \, m \)

We get:
\[ 6 - V_{0.2} = \frac{(6.6 \times 10^{-34}) \cdot (3 \times 10^{8})}{(1.6 \times 10^{-19})} \left( \frac{1}{400 \times 10^{-9}} - \frac{1}{600 \times 10^{-9}} \right) \]

### Step 5: Calculate the right-hand side

Calculating the right-hand side:
1. Calculate \( \frac{h \cdot c}{e} \):
\[ \frac{(6.6 \times 10^{-34}) \cdot (3 \times 10^{8})}{(1.6 \times 10^{-19})} \approx 1.237 \times 10^{-6} \, J \]

2. Calculate \( \frac{1}{400 \times 10^{-9}} - \frac{1}{600 \times 10^{-9}} \):
\[ \frac{1}{400 \times 10^{-9}} = 2.5 \times 10^{6} \, m^{-1} \]
\[ \frac{1}{600 \times 10^{-9}} \approx 1.6667 \times 10^{6} \, m^{-1} \]
\[ \frac{1}{400 \times 10^{-9}} - \frac{1}{600 \times 10^{-9}} \approx 0.8333 \times 10^{6} \, m^{-1} \]

3. Multiply:
\[ 1.237 \times 10^{-6} \cdot 0.8333 \times 10^{6} \approx 1.031 \, V \]

### Step 6: Solve for \( V_{0.2} \)

Now substituting back:
\[ 6 - V_{0.2} \approx 1.031 \]
\[ V_{0.2} \approx 6 - 1.031 \approx 4.969 \, V \]

### Final Answer

Thus, the new stopping potential \( V_{0.2} \) when the wavelength is 600 nm is approximately **4.97 V**.

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