In an experiment on the photoelectric effect , the slope of the cut - off voltage versus frequency of incident light is found to be `4.12xx10^(-15) V` s . The value of Planck's constant is

Options

A. `6.592xx10^(-34)` J s
B. `6.592xx10^(-31)` J s
C. `9.592xx10^(-34)` J s
D. `6.592xx10^(-30)` J s

Answer

To find the value of Planck's constant (h) using the slope of the cut-off voltage versus frequency in the photoelectric effect experiment, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the relationship**: In the photoelectric effect, the maximum kinetic energy (KE_max) of the emitted electrons is given by the equation:
\[ KE_{max} = h\nu - \phi \]
where \( \nu \) is the frequency of the incident light, \( h \) is Planck's constant, and \( \phi \) is the work function of the material.

2. **Relate cut-off voltage to energy**: The cut-off voltage (\( V_0 \)) can be expressed in terms of kinetic energy:
\[ KE_{max} = eV_0 \]
where \( e \) is the charge of an electron. Therefore, we can rewrite the equation as:
\[ eV_0 = h\nu - \phi \]

3. **Rearranging the equation**: Rearranging the above equation gives:
\[ V_0 = \frac{h}{e} \nu - \frac{\phi}{e} \]
This is in the form of a linear equation \( y = mx + c \), where:
- \( y = V_0 \) (cut-off voltage)
- \( x = \nu \) (frequency)
- \( m = \frac{h}{e} \) (slope)
- \( c = -\frac{\phi}{e} \) (y-intercept)

4. **Identify the slope**: From the problem, we know that the slope \( m \) is given as:
\[ m = 4.12 \times 10^{-15} \, \text{V s} \]
Thus, we have:
\[ \frac{h}{e} = 4.12 \times 10^{-15} \, \text{V s} \]

5. **Calculate Planck's constant**: To find \( h \), we need to multiply the slope by the charge of the electron \( e \):
\[ h = m \cdot e \]
The charge of the electron \( e \) is approximately:
\[ e = 1.6 \times 10^{-19} \, \text{C} \]
Therefore:
\[ h = (4.12 \times 10^{-15} \, \text{V s}) \times (1.6 \times 10^{-19} \, \text{C}) \]

6. **Perform the multiplication**:
\[ h = 4.12 \times 1.6 \times 10^{-15} \times 10^{-19} = 6.592 \times 10^{-34} \, \text{J s} \]

### Final Result:
The value of Planck's constant \( h \) is:
\[ h \approx 6.592 \times 10^{-34} \, \text{J s} \]