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. `sqrt(2):1`
B. `sqrt(4):1`
C. `sqrt(6):1`
D. `sqrt(8):1`

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 (\( \lambda \)) of a particle is given by the formula:
\[ \lambda = \frac{h}{mv} \]
where \( h \) is Planck's constant, \( m \) is the mass of the particle, and \( v \) is its velocity.

### Step 2: Relate velocity to kinetic energy
When a charged particle is accelerated through a potential difference \( V \), its kinetic energy (\( K \)) can be expressed as:
\[ K = QV \]
where \( Q \) is the charge of the particle. The kinetic energy can also be related to its mass and velocity:
\[ K = \frac{1}{2} mv^2 \]
Equating the two expressions for kinetic energy gives:
\[ QV = \frac{1}{2} mv^2 \implies v = \sqrt{\frac{2QV}{m}} \]

### Step 3: Substitute velocity into the de-Broglie wavelength formula
Substituting the expression for \( v \) into the de-Broglie wavelength formula:
\[ \lambda = \frac{h}{m \sqrt{\frac{2QV}{m}}} = \frac{h}{\sqrt{2mQV}} \]

### Step 4: Write expressions for the proton and alpha particle
For the proton:
\[ \lambda_p = \frac{h}{\sqrt{2m_pQ_pV}} \]
For the alpha particle (which has a charge of \( 2Q_p \) and mass \( 4m_p \)):
\[ \lambda_{\alpha} = \frac{h}{\sqrt{2m_{\alpha}Q_{\alpha}V}} = \frac{h}{\sqrt{2(4m_p)(2Q_p)V}} = \frac{h}{\sqrt{16m_pQ_pV}} \]

### Step 5: Find the ratio of the wavelengths
Now, we can find the ratio of the de-Broglie wavelengths:
\[ \frac{\lambda_p}{\lambda_{\alpha}} = \frac{\frac{h}{\sqrt{2m_pQ_pV}}}{\frac{h}{\sqrt{16m_pQ_pV}}} = \frac{\sqrt{16m_pQ_pV}}{\sqrt{2m_pQ_pV}} = \frac{4}{\sqrt{2}} = 2\sqrt{2} \]

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