A physicist was performing experiments to study the effect of varying voltage on the velocity and wavelength of the electrons. In one case, the electron was accelerated through a potential difference of 1kV and in the second case, it was accelerated through a potential difference of 2kV
In order to have half the velocity in the second case than in the first case, the potential applied should be

Options

A. 0.5 kV
B. 2kV
C. 0.25 kV
D. 0.75 kV

Explanation

To solve the problem, we need to analyze the relationship between the potential difference (voltage), kinetic energy, and the velocity of the electrons. Here’s a step-by-step breakdown of the solution:

### Step 1: Understand the Kinetic Energy of the Electron
The kinetic energy (KE) of an electron accelerated through a potential difference \( V \) is given by the formula:
\[ KE = eV \]
where \( e \) is the charge of the electron and \( V \) is the potential difference.

### Step 2: Set Up the Kinetic Energy Equations
For the first case where the electron is accelerated through a potential difference of 1 kV (or 1000 V):
\[ KE_1 = e \cdot 1000 \]
For the second case where the electron is accelerated through a potential difference of \( V_2 \) (which we will denote as \( x \)):
\[ KE_2 = e \cdot x \]

### Step 3: Relate Kinetic Energy to Velocity
The kinetic energy can also be expressed in terms of the mass \( m \) of the electron and its velocity \( v \):
\[ KE = \frac{1}{2} mv^2 \]
Thus, we can write:
\[ \frac{1}{2} mv_1^2 = e \cdot 1000 \quad \text{(for the first case)} \]
\[ \frac{1}{2} mv_2^2 = e \cdot x \quad \text{(for the second case)} \]

### Step 4: Express the Velocity Relationship
According to the problem, the velocity in the second case \( v_2 \) is half of the velocity in the first case \( v_1 \):
\[ v_2 = \frac{v_1}{2} \]

### Step 5: Substitute \( v_2 \) into the Kinetic Energy Equation
Substituting \( v_2 \) into the kinetic energy equation for the second case:
\[ \frac{1}{2} m \left(\frac{v_1}{2}\right)^2 = e \cdot x \]
This simplifies to:
\[ \frac{1}{2} m \cdot \frac{v_1^2}{4} = e \cdot x \]
\[ \frac{1}{8} mv_1^2 = e \cdot x \]

### Step 6: Relate the Two Kinetic Energy Equations
Now we can relate the two kinetic energy equations:
From the first case, we have:
\[ \frac{1}{2} mv_1^2 = e \cdot 1000 \]
Substituting \( \frac{1}{2} mv_1^2 \) from the first case into the equation for the second case:
\[ \frac{1}{8} mv_1^2 = e \cdot x \implies \frac{1}{8} \cdot (2 \cdot e \cdot 1000) = e \cdot x \]
This simplifies to:
\[ \frac{1000}{4} = x \]
\[ x = 250 \text{ volts} \]

### Step 7: Convert to Kilovolts
Since the question asks for the potential in kilovolts:
\[ x = \frac{250}{1000} = 0.25 \text{ kV} \]

### Final Answer
The potential applied in the second case should be **0.25 kV**.

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