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 wavelength of a proton to that of an alpha particle, both accelerated through the same potential difference, we can follow these steps:

### Step 1: Understand the de Broglie wavelength formula
The de Broglie wavelength (λ) 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
The kinetic energy (KE) of a particle accelerated through a potential difference \(V\) is given by:
\[ KE = qV \]
where \(q\) is the charge of the particle. For a proton and an alpha particle, we can express their kinetic energies as:
\[ KE_p = q_p V \quad \text{(for proton)} \]
\[ KE_{\alpha} = q_{\alpha} V \quad \text{(for alpha particle)} \]

### Step 3: Express momentum in terms of kinetic energy
The momentum \(p\) can be expressed using the kinetic energy:
\[ KE = \frac{p^2}{2m} \implies p = \sqrt{2m \cdot KE} \]
Thus, we can write:
\[ p_p = \sqrt{2m_p \cdot KE_p} = \sqrt{2m_p \cdot q_p V} \]
\[ p_{\alpha} = \sqrt{2m_{\alpha} \cdot KE_{\alpha}} = \sqrt{2m_{\alpha} \cdot q_{\alpha} V} \]

### Step 4: Substitute momentum into the de Broglie wavelength formula
Now substituting the expressions for momentum into the de Broglie wavelength formula:
\[ \lambda_p = \frac{h}{p_p} = \frac{h}{\sqrt{2m_p \cdot q_p V}} \]
\[ \lambda_{\alpha} = \frac{h}{p_{\alpha}} = \frac{h}{\sqrt{2m_{\alpha} \cdot q_{\alpha} V}} \]

### Step 5: Find the ratio of the wavelengths
Now we can find the ratio of the de Broglie wavelengths:
\[ \frac{\lambda_p}{\lambda_{\alpha}} = \frac{\frac{h}{\sqrt{2m_p \cdot q_p V}}}{\frac{h}{\sqrt{2m_{\alpha} \cdot q_{\alpha} V}}} = \frac{\sqrt{2m_{\alpha} \cdot q_{\alpha} V}}{\sqrt{2m_p \cdot q_p V}} \]
The \(h\) and \(V\) cancel out:
\[ \frac{\lambda_p}{\lambda_{\alpha}} = \sqrt{\frac{m_{\alpha} \cdot q_{\alpha}}{m_p \cdot q_p}} \]

### Step 6: Substitute known values
For the proton:
- Mass \(m_p = m\)
- Charge \(q_p = e\)

For the alpha particle:
- Mass \(m_{\alpha} = 4m\) (since an alpha particle consists of 2 protons and 2 neutrons)
- Charge \(q_{\alpha} = 2e\)

Substituting these values into the ratio:
\[ \frac{\lambda_p}{\lambda_{\alpha}} = \sqrt{\frac{(4m)(2e)}{m(e)}} = \sqrt{\frac{8m \cdot e}{m \cdot e}} = \sqrt{8} = 2\sqrt{2} \]

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