If the stationary proton and α-particle are accelerated through same potential difference, the ratio of debroglie’s wavelengths will be :

Options

A. 2
B. 1
C. `2sqrt2`
D. 0.25

Answer

To solve the problem of finding the ratio of the de Broglie wavelengths of a proton and an alpha particle when both are accelerated through the same potential difference, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the de Broglie Wavelength Formula**:
The de Broglie wavelength (\( \lambda \)) is given by the formula:
\[ \lambda = \frac{h}{p} \]
where \( h \) is Planck's constant and \( p \) is the momentum of the particle.

2. **Relate Momentum to Kinetic Energy**:
The momentum \( p \) can be expressed in terms of kinetic energy (\( KE \)):
\[ KE = \frac{p^2}{2m} \implies p = \sqrt{2m \cdot KE} \]
Therefore, we can rewrite the de Broglie wavelength as:
\[ \lambda = \frac{h}{\sqrt{2m \cdot KE}} \]

3. **Calculate Kinetic Energy for Each Particle**:
When a charged particle is accelerated through a potential difference \( V \), its kinetic energy is given by:
\[ KE = QV \]
where \( Q \) is the charge of the particle.

- For the proton, with charge \( e \):
\[ KE_p = eV \]
- For the alpha particle, with charge \( 2e \):
\[ KE_{\alpha} = 2eV \]

4. **Substitute Kinetic Energy into the Wavelength Formula**:
Now we can express the de Broglie wavelengths for both particles:
- For the proton:
\[ \lambda_p = \frac{h}{\sqrt{2m \cdot eV}} \]
- For the alpha particle (mass \( 4m \)):
\[ \lambda_{\alpha} = \frac{h}{\sqrt{2(4m) \cdot (2eV)}} = \frac{h}{\sqrt{8m \cdot eV}} \]

5. **Calculate the Ratio of the Wavelengths**:
Now we can find the ratio of the de Broglie wavelengths:
\[ \frac{\lambda_p}{\lambda_{\alpha}} = \frac{\frac{h}{\sqrt{2m \cdot eV}}}{\frac{h}{\sqrt{8m \cdot eV}}} \]
Simplifying this gives:
\[ \frac{\lambda_p}{\lambda_{\alpha}} = \frac{\sqrt{8m \cdot eV}}{\sqrt{2m \cdot eV}} = \frac{\sqrt{8}}{\sqrt{2}} = \sqrt{4} = 2 \]

6. **Final Ratio**:
Therefore, the ratio of the de Broglie wavelengths of the proton to the alpha particle is:
\[ \frac{\lambda_p}{\lambda_{\alpha}} = 2 \]

### Conclusion:
The ratio of the de Broglie wavelengths of the proton to the alpha particle is \( 2:1 \).

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