If maximum velocity with which an electron can be emitted is `4xx10^8` `cm s^-1` , then find the stopping potential `V_o` (mass of electron`=9xx10^(-31)` kg)

Options

A. 30 volt
B. 45 volt
C. 59 volt
D. Information is insufficient

Answer

To find the stopping potential \( V_0 \) when the maximum velocity of an emitted electron is given, we can follow these steps:

### Step 1: Convert the velocity to meters per second
The maximum velocity \( V \) is given as \( 4 \times 10^8 \) cm/s. We need to convert this to meters per second:

\[ V = 4 \times 10^8 \, \text{cm/s} = 4 \times 10^6 \, \text{m/s} \]

### Step 2: Write the formula for maximum kinetic energy
The maximum kinetic energy \( K_{\text{max}} \) of the electron can be expressed using the formula:

\[ K_{\text{max}} = \frac{1}{2} m V^2 \]

where \( m \) is the mass of the electron and \( V \) is its velocity.

### Step 3: Relate kinetic energy to stopping potential
The stopping potential \( V_0 \) is related to the kinetic energy by the equation:

\[ K_{\text{max}} = e V_0 \]

where \( e \) is the charge of the electron (\( 1.6 \times 10^{-19} \) C).

### Step 4: Combine the equations
From the two equations, we can set them equal to each other:

\[ \frac{1}{2} m V^2 = e V_0 \]

Rearranging gives us:

\[ V_0 = \frac{1}{2} \frac{m V^2}{e} \]

### Step 5: Substitute the known values
Now we can substitute the known values into the equation. The mass of the electron \( m = 9 \times 10^{-31} \) kg, the velocity \( V = 4 \times 10^6 \) m/s, and the charge \( e = 1.6 \times 10^{-19} \) C:

\[ V_0 = \frac{1}{2} \frac{(9 \times 10^{-31} \, \text{kg}) (4 \times 10^6 \, \text{m/s})^2}{1.6 \times 10^{-19} \, \text{C}} \]

### Step 6: Calculate \( V_0 \)
First, calculate \( (4 \times 10^6)^2 \):

\[ (4 \times 10^6)^2 = 16 \times 10^{12} \, \text{m}^2/\text{s}^2 \]

Now substitute this back into the equation:

\[ V_0 = \frac{1}{2} \frac{(9 \times 10^{-31}) (16 \times 10^{12})}{1.6 \times 10^{-19}} \]

Calculating the numerator:

\[ 9 \times 10^{-31} \times 16 \times 10^{12} = 144 \times 10^{-19} \, \text{kg m}^2/\text{s}^2 \]

Now substituting this into the equation for \( V_0 \):

\[ V_0 = \frac{1}{2} \frac{144 \times 10^{-19}}{1.6 \times 10^{-19}} = \frac{72}{1.6} = 45 \, \text{volts} \]

### Final Answer
Thus, the stopping potential \( V_0 \) is:

\[ \boxed{45 \, \text{volts}} \]