An electron accelerated through a potential difference `V_(1)` has a de-Broglie wavelength of `lambda`.When the potential is changed to `V_(2)`,its de-Broglie wavelength increases by `50%`.The value of `(V_(1)V_(2))` is equal to

Options

A. `4`
B. `9/4`
C. `3`
D. `3/2`

Answer

To solve the problem, we need to analyze the relationship between the de-Broglie wavelength of an electron and the potential difference through which it is accelerated.

### Step-by-Step Solution:

1. **Understanding de-Broglie Wavelength**:
The de-Broglie wavelength (\( \lambda \)) of a particle is given by the formula:
\[ \lambda = \frac{h}{p} \]
where \( h \) is Planck's constant and \( p \) is the momentum of the particle. For an electron accelerated through a potential difference \( V \), the momentum can be expressed in terms of kinetic energy.

2. **Kinetic Energy and Momentum**:
The kinetic energy (\( K.E. \)) gained by the electron when accelerated through a potential \( V \) is:
\[ K.E. = eV \]
where \( e \) is the charge of the electron. The momentum \( p \) can be expressed as:
\[ p = \sqrt{2mK.E.} = \sqrt{2meV} \]
where \( m \) is the mass of the electron.

3. **Substituting into the Wavelength Formula**:
Substituting the expression for momentum into the de-Broglie wavelength formula gives:
\[ \lambda = \frac{h}{\sqrt{2meV}} \]

4. **Wavelengths at Different Potentials**:
Let \( \lambda_1 \) be the wavelength when the potential is \( V_1 \) and \( \lambda_2 \) when the potential is \( V_2 \). Thus, we have:
\[ \lambda_1 = \frac{h}{\sqrt{2meV_1}} \quad \text{and} \quad \lambda_2 = \frac{h}{\sqrt{2meV_2}} \]

5. **Relationship Between Wavelengths**:
According to the problem, when the potential changes from \( V_1 \) to \( V_2 \), the wavelength increases by 50%. Therefore:
\[ \lambda_2 = 1.5 \lambda_1 \]

6. **Setting Up the Equation**:
Substituting the expressions for \( \lambda_1 \) and \( \lambda_2 \):
\[ \frac{h}{\sqrt{2meV_2}} = 1.5 \cdot \frac{h}{\sqrt{2meV_1}} \]
Cancelling \( h \) and \( \sqrt{2me} \) from both sides gives:
\[ \frac{1}{\sqrt{V_2}} = 1.5 \cdot \frac{1}{\sqrt{V_1}} \]

7. **Cross-Multiplying**:
Cross-multiplying yields:
\[ \sqrt{V_1} = 1.5 \sqrt{V_2} \]

8. **Squaring Both Sides**:
Squaring both sides results in:
\[ V_1 = (1.5)^2 V_2 = \frac{9}{4} V_2 \]

9. **Finding the Ratio**:
Rearranging gives:
\[ \frac{V_1}{V_2} = \frac{9}{4} \]

### Final Answer:
The value of \( V_1 V_2 \) can be expressed as:
\[ \frac{V_1}{V_2} = \frac{9}{4} \]