Maximum velocity of photo electrons is `3.5 xx 10^(6) m//s`. If the specific charge of an electron is `1.75 xx 10^(11)` C/kg, stopping potential of the electron is

Options

A. 7 volt
B. 3.5 volt
C. 35 volt
D. 10.5 volt

Explanation

To find the stopping potential of the electron, we can use the relationship between kinetic energy and electric potential energy. Let's break down the steps:

### Step-by-Step Solution:

1. **Understanding the Relationship**:
The kinetic energy (KE) of the photoelectron can be expressed as:
\[ KE = \frac{1}{2} mv^2 \]
where \( m \) is the mass of the electron and \( v \) is the maximum velocity of the photoelectron.

2. **Relating Kinetic Energy to Stopping Potential**:
The stopping potential \( V \) is related to the kinetic energy by:
\[ KE = qV \]
where \( q \) is the charge of the electron.

3. **Substituting for Kinetic Energy**:
From the above two equations, we can equate the kinetic energy to the charge times the stopping potential:
\[ \frac{1}{2} mv^2 = qV \]

4. **Specific Charge**:
The specific charge of the electron is given as:
\[ \frac{q}{m} = 1.75 \times 10^{11} \, \text{C/kg} \]
From this, we can express \( q \) in terms of \( m \):
\[ q = \left(1.75 \times 10^{11}\right) m \]

5. **Substituting Specific Charge into the Equation**:
Rearranging the kinetic energy equation gives:
\[ V = \frac{\frac{1}{2} mv^2}{q} \]
Substituting \( q \) from the specific charge:
\[ V = \frac{\frac{1}{2} mv^2}{\left(1.75 \times 10^{11}\right) m} \]
The mass \( m \) cancels out:
\[ V = \frac{1}{2} \cdot \frac{v^2}{1.75 \times 10^{11}} \]

6. **Substituting the Maximum Velocity**:
Now, substitute the maximum velocity \( v = 3.5 \times 10^6 \, \text{m/s} \):
\[ V = \frac{1}{2} \cdot \frac{(3.5 \times 10^6)^2}{1.75 \times 10^{11}} \]

7. **Calculating the Value**:
First, calculate \( (3.5 \times 10^6)^2 \):
\[ (3.5 \times 10^6)^2 = 12.25 \times 10^{12} = 1.225 \times 10^{13} \]
Now substitute this back into the equation for \( V \):
\[ V = \frac{1}{2} \cdot \frac{1.225 \times 10^{13}}{1.75 \times 10^{11}} \]
Simplifying this:
\[ V = \frac{1.225 \times 10^{13}}{3.5 \times 10^{11}} = 35 \, \text{volts} \]

### Final Answer:
The stopping potential \( V \) of the electron is **35 volts**.

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