In a photoelectric experiment it was found that the stopping potential decreases from 1.85 V to 0.82 V as the wavelength of the incident light is varied from 300 nm to 400 nm. Calculate the value of the Planck constant from these data.

Options

A. `4.12 xx 10^(-15)` eVs
B. `2.5 xx 10^(10) eVs`
C. `5.2 xx 10^(-15)` eVs
D. `7.5 xx 10^(-8)` eVs

Explanation

To calculate the value of the Planck constant (h) using the data from the photoelectric experiment, we can follow these steps:

### Step 1: Understand the photoelectric effect
The photoelectric effect states that when light of a certain frequency shines on a metal surface, it can eject electrons from that surface. The maximum kinetic energy (Kmax) of the emitted electrons can be expressed as:
\[ K_{max} = h \frac{c}{\lambda} - \phi \]
where:
- \( K_{max} \) is the maximum kinetic energy of the emitted electrons,
- \( h \) is the Planck constant,
- \( c \) is the speed of light (approximately \( 3 \times 10^8 \) m/s),
- \( \lambda \) is the wavelength of the incident light,
- \( \phi \) is the work function of the material.

### Step 2: Relate stopping potential to kinetic energy
The stopping potential (V) is related to the maximum kinetic energy by:
\[ K_{max} = eV \]
where \( e \) is the charge of an electron (approximately \( 1.6 \times 10^{-19} \) C). Therefore, we can express the equation for stopping potential as:
\[ eV = h \frac{c}{\lambda} - \phi \]

### Step 3: Set up equations for two wavelengths
We have two different stopping potentials and wavelengths:
1. For \( \lambda_1 = 300 \) nm (or \( 300 \times 10^{-9} \) m), \( V_1 = 1.85 \) V
2. For \( \lambda_2 = 400 \) nm (or \( 400 \times 10^{-9} \) m), \( V_2 = 0.82 \) V

From the stopping potential equations, we can write:
1. \( eV_1 = h \frac{c}{\lambda_1} - \phi \)
2. \( eV_2 = h \frac{c}{\lambda_2} - \phi \)

### Step 4: Subtract the equations
By subtracting the second equation from the first, we eliminate \( \phi \):
\[ eV_1 - eV_2 = h \left( \frac{c}{\lambda_1} - \frac{c}{\lambda_2} \right) \]

### Step 5: Substitute known values
Substituting the values:
- \( V_1 = 1.85 \) V
- \( V_2 = 0.82 \) V
- \( \lambda_1 = 300 \times 10^{-9} \) m
- \( \lambda_2 = 400 \times 10^{-9} \) m
- \( e = 1.6 \times 10^{-19} \) C
- \( c = 3 \times 10^8 \) m/s

The equation becomes:
\[ e(1.85 - 0.82) = h \left( \frac{3 \times 10^8}{300 \times 10^{-9}} - \frac{3 \times 10^8}{400 \times 10^{-9}} \right) \]

### Step 6: Calculate the left side
Calculating the left side:
\[ e(1.85 - 0.82) = 1.6 \times 10^{-19} \times 1.03 \]
\[ = 1.646 \times 10^{-19} \text{ J} \]

### Step 7: Calculate the right side
Calculating the right side:
\[ \frac{3 \times 10^8}{300 \times 10^{-9}} = 1 \times 10^{15} \text{ Hz} \]
\[ \frac{3 \times 10^8}{400 \times 10^{-9}} = 0.75 \times 10^{15} \text{ Hz} \]
So,
\[ h \left( 1 \times 10^{15} - 0.75 \times 10^{15} \right) = h \times 0.25 \times 10^{15} \]

### Step 8: Solve for Planck's constant (h)
Now we can equate both sides:
\[ 1.646 \times 10^{-19} = h \times 0.25 \times 10^{15} \]
\[ h = \frac{1.646 \times 10^{-19}}{0.25 \times 10^{15}} \]
\[ h = \frac{1.646}{0.25} \times 10^{-4} \]
\[ h = 6.584 \times 10^{-34} \text{ J s} \]

### Final Answer
The value of the Planck constant \( h \) is approximately \( 6.584 \times 10^{-34} \text{ J s} \).

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