In one experiment , a proton having kinetic energy of 1 eV is accelerated through a potential difference of 3 V. In another experiment, an `alpha`-particle having initial kinetic energy 20 eV is retarded by a potential difference of 2 V. Calculate the ratio of de-Broglie wavelengths of proton and `alpha` - particle.

Answer

To solve the problem, we need to calculate the de-Broglie wavelengths of a proton and an alpha particle and then find the ratio of these wavelengths.

### Step 1: Calculate the total kinetic energy of the proton
The total kinetic energy (KE) of the proton is given by the sum of its initial kinetic energy and the energy gained from being accelerated through a potential difference.

- Initial kinetic energy of the proton = 1 eV
- Potential difference = 3 V

Using the formula:
\[ KE_{\text{proton}} = KE_{\text{initial}} + qV \]
Where \( q \) is the charge of the proton (1 eV corresponds to 1 unit of charge).

Thus,
\[ KE_{\text{proton}} = 1 \, \text{eV} + 3 \, \text{eV} = 4 \, \text{eV} \]

### Step 2: Calculate the total kinetic energy of the alpha particle
The total kinetic energy of the alpha particle is calculated similarly, but since it is being retarded by a potential difference, we subtract the energy lost.

- Initial kinetic energy of the alpha particle = 20 eV
- Potential difference = 2 V

Using the formula:
\[ KE_{\text{alpha}} = KE_{\text{initial}} - qV \]
Where \( q \) for an alpha particle is 2 (since it has a charge of +2e).

Thus,
\[ KE_{\text{alpha}} = 20 \, \text{eV} - 2 \, \text{eV} = 18 \, \text{eV} \]

### Step 3: Use the de-Broglie wavelength formula
The de-Broglie wavelength \( \lambda \) is given by the formula:
\[ \lambda = \frac{h}{\sqrt{2m \cdot KE}} \]
Where:
- \( h \) is Planck's constant,
- \( m \) is the mass of the particle,
- \( KE \) is the kinetic energy.

### Step 4: Calculate the ratio of the de-Broglie wavelengths
We need to find the ratio of the de-Broglie wavelengths of the proton and the alpha particle:
\[ \frac{\lambda_{\text{proton}}}{\lambda_{\text{alpha}}} = \frac{\frac{h}{\sqrt{2m_{\text{proton}} \cdot KE_{\text{proton}}}}}{\frac{h}{\sqrt{2m_{\text{alpha}} \cdot KE_{\text{alpha}}}}} \]

This simplifies to:
\[ \frac{\lambda_{\text{proton}}}{\lambda_{\text{alpha}}} = \sqrt{\frac{m_{\text{alpha}} \cdot KE_{\text{alpha}}}{m_{\text{proton}} \cdot KE_{\text{proton}}}} \]

### Step 5: Substitute the values
- Mass of proton \( m_{\text{proton}} = 1 \) (arbitrary unit)
- Mass of alpha particle \( m_{\text{alpha}} = 4 \) (since an alpha particle consists of 2 protons and 2 neutrons)
- \( KE_{\text{proton}} = 4 \, \text{eV} \)
- \( KE_{\text{alpha}} = 18 \, \text{eV} \)

Now substituting these values:
\[ \frac{\lambda_{\text{proton}}}{\lambda_{\text{alpha}}} = \sqrt{\frac{4 \cdot 18}{1 \cdot 4}} = \sqrt{\frac{72}{4}} = \sqrt{18} = 3\sqrt{2} \]

### Final Ratio
Thus, the ratio of the de-Broglie wavelengths of the proton to the alpha particle is:
\[ \frac{\lambda_{\text{proton}}}{\lambda_{\text{alpha}}} = 3\sqrt{2} \]