Find the ratio of de Broglie wavelength of a proton and as `alpha`-particle which have been accelerated through same potential difference.

Options

A. `2sqrt2:1`
B. `3:2`
C. `3sqrt2:1`
D. `2:1`

Explanation

To find the ratio of the de Broglie wavelengths of a proton and an alpha particle that have been accelerated through the same potential difference, we can follow these steps:

### Step 1: Understand the Kinetic Energy
When a charged particle is accelerated through a potential difference \( V \), it gains kinetic energy given by:
\[ KE = qV \]
where \( q \) is the charge of the particle.

### Step 2: Relate Kinetic Energy to Velocity
The kinetic energy can also be expressed in terms of mass \( m \) and velocity \( v \):
\[ KE = \frac{1}{2} mv^2 \]
Setting these two expressions for kinetic energy equal gives us:
\[ \frac{1}{2} mv^2 = qV \]
From this, we can solve for the velocity \( v \):
\[ v = \sqrt{\frac{2qV}{m}} \]

### Step 3: Calculate Momentum
The momentum \( p \) of the particle is given by:
\[ p = mv \]
Substituting our expression for \( v \) into this equation, we get:
\[ p = m \sqrt{\frac{2qV}{m}} = \sqrt{2mqV} \]

### Step 4: Write the de Broglie Wavelength
The de Broglie wavelength \( \lambda \) is given by:
\[ \lambda = \frac{h}{p} \]
where \( h \) is Planck's constant. Therefore, for the proton and alpha particle, we have:
\[ \lambda_p = \frac{h}{\sqrt{2m_pq_pV}} \quad \text{(for proton)} \]
\[ \lambda_{\alpha} = \frac{h}{\sqrt{2m_{\alpha}q_{\alpha}V}} \quad \text{(for alpha particle)} \]

### Step 5: Find the Ratio of Wavelengths
Now, we can find the ratio of the de Broglie wavelengths:
\[ \frac{\lambda_p}{\lambda_{\alpha}} = \frac{\sqrt{2m_{\alpha}q_{\alpha}V}}{\sqrt{2m_pq_pV}} \]
The \( V \) cancels out, and we have:
\[ \frac{\lambda_p}{\lambda_{\alpha}} = \sqrt{\frac{m_{\alpha}q_{\alpha}}{m_pq_p}} \]

### Step 6: Substitute Known Values
For an alpha particle:
- Mass \( m_{\alpha} = 4m_p \) (since an alpha particle consists of 2 protons and 2 neutrons)
- Charge \( q_{\alpha} = 2q_p \)

Substituting these values into the ratio gives:
\[ \frac{\lambda_p}{\lambda_{\alpha}} = \sqrt{\frac{4m_p \cdot 2q_p}{m_p \cdot q_p}} = \sqrt{\frac{8m_pq_p}{m_pq_p}} = \sqrt{8} = 2\sqrt{2} \]

### Final Answer
Thus, the ratio of the de Broglie wavelength of a proton to that of an alpha particle is:
\[ \frac{\lambda_p}{\lambda_{\alpha}} = 2\sqrt{2} : 1 \]