A proton and an alpha - particle are accelerated through same potential difference. Then, the ratio of de-Broglie wavelength of proton and alpha-particle is

Options

A. `sqrt2`
B. `(1)/(sqrt2)`
C. `2sqrt2`
D. None of these

Explanation

To find 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 1: Understand the de-Broglie wavelength formula
The de-Broglie wavelength (λ) 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.

### Step 2: Relate momentum to kinetic energy
When a charged particle is accelerated through a potential difference \( V \), it gains kinetic energy equal to the work done on it by the electric field:
\[ KE = qV \]
where \( q \) is the charge of the particle. The kinetic energy can also be expressed in terms of momentum:
\[ KE = \frac{p^2}{2m} \]
Equating the two expressions gives:
\[ qV = \frac{p^2}{2m} \]
From this, we can express momentum \( p \):
\[ p = \sqrt{2mqV} \]

### Step 3: Calculate the de-Broglie wavelength for both particles
For the proton:
- Charge of proton \( q_p = e \)
- Mass of proton \( m_p = m \)

The de-Broglie wavelength of the proton is:
\[ \lambda_p = \frac{h}{p_p} = \frac{h}{\sqrt{2m_p q_p V}} = \frac{h}{\sqrt{2m e V}} \]

For the alpha particle:
- Charge of alpha particle \( q_{\alpha} = 2e \)
- Mass of alpha particle \( m_{\alpha} = 4m \)

The de-Broglie wavelength of the alpha particle is:
\[ \lambda_{\alpha} = \frac{h}{p_{\alpha}} = \frac{h}{\sqrt{2m_{\alpha} q_{\alpha} V}} = \frac{h}{\sqrt{2(4m)(2e)V}} = \frac{h}{\sqrt{16m e V}} = \frac{h}{4\sqrt{m e V}} \]

### Step 4: Find the ratio of the wavelengths
Now, we can find the ratio of the de-Broglie wavelengths:
\[ \frac{\lambda_p}{\lambda_{\alpha}} = \frac{\frac{h}{\sqrt{2m e V}}}{\frac{h}{4\sqrt{m e V}}} \]
This simplifies to:
\[ \frac{\lambda_p}{\lambda_{\alpha}} = \frac{4\sqrt{m e V}}{\sqrt{2m e V}} = \frac{4}{\sqrt{2}} = 2\sqrt{2} \]

### Final Answer
The ratio of the de-Broglie wavelength of the proton to that of the alpha particle is:
\[ \frac{\lambda_p}{\lambda_{\alpha}} = 2\sqrt{2} \]