Light of wavelength `0.6 mum `from a sodium lamp falls on a photocell and causes the emission of photoelectrons for which the stopping potential is 0.5 V. With wavelength `0.4 mum` from a sodium lamp, the stopping potential is 1.5 V. With this data , the value of h/e is

Options

A. `4xx10^(-59)Vs`
B. `0.25xx10^(-15)Vs`
C. `4xx10^(-15)Vs`
D. `4xx10^(-8)Vs`.

Explanation

To find the value of \( \frac{h}{e} \) using the given data, we can follow these steps:

### Step 1: Write the energy equations for both wavelengths
The energy of the photons can be expressed as:
\[ E = \frac{hc}{\lambda} \]
For two different wavelengths \( \lambda_1 = 0.6 \, \mu m \) and \( \lambda_2 = 0.4 \, \mu m \), we can write:
1. For \( \lambda_1 \):
\[ \frac{hc}{\lambda_1} - W_0 = eV_1 \]
2. For \( \lambda_2 \):
\[ \frac{hc}{\lambda_2} - W_0 = eV_2 \]

### Step 2: Rearrange the equations
Rearranging both equations gives:
1. \( \frac{hc}{\lambda_1} = W_0 + eV_1 \)
2. \( \frac{hc}{\lambda_2} = W_0 + eV_2 \)

### Step 3: Subtract the two equations
Subtracting the first equation from the second:
\[ \frac{hc}{\lambda_2} - \frac{hc}{\lambda_1} = eV_2 - eV_1 \]
Factoring out \( hc \):
\[ hc \left( \frac{1}{\lambda_2} - \frac{1}{\lambda_1} \right) = e(V_2 - V_1) \]

### Step 4: Substitute known values
Substituting the known values:
- \( \lambda_1 = 0.6 \, \mu m = 0.6 \times 10^{-6} \, m \)
- \( \lambda_2 = 0.4 \, \mu m = 0.4 \times 10^{-6} \, m \)
- \( V_1 = 0.5 \, V \)
- \( V_2 = 1.5 \, V \)

Calculating \( V_2 - V_1 \):
\[ V_2 - V_1 = 1.5 - 0.5 = 1.0 \, V \]

### Step 5: Calculate \( \frac{1}{\lambda_2} - \frac{1}{\lambda_1} \)
Calculating \( \frac{1}{\lambda_2} - \frac{1}{\lambda_1} \):
\[ \frac{1}{\lambda_2} = \frac{1}{0.4 \times 10^{-6}} = 2.5 \times 10^{6} \, m^{-1} \]
\[ \frac{1}{\lambda_1} = \frac{1}{0.6 \times 10^{-6}} \approx 1.6667 \times 10^{6} \, m^{-1} \]
\[ \frac{1}{\lambda_2} - \frac{1}{\lambda_1} = 2.5 \times 10^{6} - 1.6667 \times 10^{6} = 0.8333 \times 10^{6} \, m^{-1} \]

### Step 6: Substitute back to find \( \frac{h}{e} \)
Now we substitute back into the equation:
\[ hc \left( 0.8333 \times 10^{6} \right) = e(1.0) \]
Thus,
\[ \frac{h}{e} = \frac{1}{c \left( 0.8333 \times 10^{6} \right)} \]

Using \( c \approx 3 \times 10^{8} \, m/s \):
\[ \frac{h}{e} = \frac{1}{(3 \times 10^{8}) \times (0.8333 \times 10^{6})} \]
Calculating this gives:
\[ \frac{h}{e} \approx \frac{1}{2.5 \times 10^{14}} \approx 4 \times 10^{-15} \, V \cdot s \]

### Final Result
Thus, the value of \( \frac{h}{e} \) is:
\[ \frac{h}{e} \approx 4 \times 10^{-15} \, V \cdot s \]

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