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. `10xx10^(-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 down Einstein's photoelectric equation
Einstein's photoelectric equation is given by:
\[ E \cdot V_0 = \frac{hc}{\lambda} - W \]
where:
- \( E \) is the charge of the electron (denoted as \( e \)),
- \( V_0 \) is the stopping potential,
- \( h \) is Planck's constant,
- \( c \) is the speed of light,
- \( \lambda \) is the wavelength of light,
- \( W \) is the work function of the metal.

### Step 2: Set up equations for both cases
For the first case with \( \lambda_1 = 0.6 \, \mu m \) and \( V_{01} = 0.5 \, V \):
\[ 0.5e = \frac{hc}{0.6 \times 10^{-6}} - W \quad \text{(Equation 1)} \]

For the second case with \( \lambda_2 = 0.4 \, \mu m \) and \( V_{02} = 1.5 \, V \):
\[ 1.5e = \frac{hc}{0.4 \times 10^{-6}} - W \quad \text{(Equation 2)} \]

### Step 3: Subtract Equation 1 from Equation 2
Now, we subtract Equation 1 from Equation 2 to eliminate \( W \):
\[ (1.5e - 0.5e) = \left(\frac{hc}{0.4 \times 10^{-6}} - W\right) - \left(\frac{hc}{0.6 \times 10^{-6}} - W\right) \]
This simplifies to:
\[ 1e = \frac{hc}{0.4 \times 10^{-6}} - \frac{hc}{0.6 \times 10^{-6}} \]

### Step 4: Factor out \( hc \)
Factoring out \( hc \):
\[ 1e = hc \left(\frac{1}{0.4 \times 10^{-6}} - \frac{1}{0.6 \times 10^{-6}}\right) \]

### Step 5: Simplify the right-hand side
Calculating the right-hand side:
\[ \frac{1}{0.4 \times 10^{-6}} - \frac{1}{0.6 \times 10^{-6}} = \frac{1}{0.4} - \frac{1}{0.6} = \frac{3}{1.2} = \frac{5}{2.4} = \frac{5 \times 10^{6}}{2.4} \]
This gives:
\[ 1e = hc \cdot \left(\frac{5}{2.4} \times 10^{6}\right) \]

### Step 6: Solve for \( \frac{h}{e} \)
Rearranging gives:
\[ \frac{h}{e} = \frac{1}{\left(\frac{5}{2.4} \times 10^{6}\right)} = \frac{2.4}{5} \times 10^{-6} \]
Calculating this gives:
\[ \frac{h}{e} = \frac{12}{15} \times 10^{-15} = 4 \times 10^{-15} \, \text{V s} \]

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

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