An `alpha`-particle and a proton are accelerated from rest through the same potential difference V. Find the ratio of de-Broglie wavelength associated with them.

Explanation

To find the ratio of the de-Broglie wavelengths associated with an alpha particle and a proton that are accelerated from rest through the same potential difference \( V \), we can follow these steps:

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

### Step 2: Write the Kinetic Energy for Alpha Particle and Proton
For the alpha particle:
\[ KE_{\alpha} = Q_{\alpha} \cdot V \]
For the proton:
\[ KE_{p} = Q_{p} \cdot V \]

### Step 3: Relate Kinetic Energy to Momentum
The kinetic energy can also be expressed in terms of momentum \( p \):
\[ KE = \frac{p^2}{2m} \]
Thus, we can write:
\[ p_{\alpha}^2 = 2m_{\alpha} KE_{\alpha} \quad \text{and} \quad p_{p}^2 = 2m_{p} KE_{p} \]

### Step 4: Substitute Kinetic Energy into Momentum Equations
Substituting the expressions for kinetic energy:
\[ p_{\alpha}^2 = 2m_{\alpha} (Q_{\alpha} \cdot V) \]
\[ p_{p}^2 = 2m_{p} (Q_{p} \cdot V) \]

### Step 5: Find the Expressions for Momentum
Taking the square root gives:
\[ p_{\alpha} = \sqrt{2m_{\alpha} Q_{\alpha} V} \]
\[ p_{p} = \sqrt{2m_{p} Q_{p} V} \]

### Step 6: Write the de-Broglie Wavelength Formula
The de-Broglie wavelength \( \lambda \) is given by:
\[ \lambda = \frac{h}{p} \]
where \( h \) is Planck's constant.

### Step 7: Write the Wavelengths for Alpha Particle and Proton
Thus, the de-Broglie wavelengths for the alpha particle and proton are:
\[ \lambda_{\alpha} = \frac{h}{p_{\alpha}} = \frac{h}{\sqrt{2m_{\alpha} Q_{\alpha} V}} \]
\[ \lambda_{p} = \frac{h}{p_{p}} = \frac{h}{\sqrt{2m_{p} Q_{p} V}} \]

### Step 8: Find the Ratio of the Wavelengths
The ratio of the de-Broglie wavelengths is:
\[ \frac{\lambda_{\alpha}}{\lambda_{p}} = \frac{\sqrt{2m_{p} Q_{p} V}}{\sqrt{2m_{\alpha} Q_{\alpha} V}} = \frac{\sqrt{m_{p} Q_{p}}}{\sqrt{m_{\alpha} Q_{\alpha}}} \]

### Step 9: 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:
\[ \frac{\lambda_{\alpha}}{\lambda_{p}} = \frac{\sqrt{m_{p} Q_{p}}}{\sqrt{4m_{p} \cdot 2Q_{p}}} = \frac{\sqrt{m_{p} Q_{p}}}{\sqrt{8m_{p} Q_{p}}} = \frac{1}{\sqrt{8}} = \frac{1}{2\sqrt{2}} \]

### Final Ratio of Wavelengths
Thus, the ratio of the de-Broglie wavelengths is:
\[ \frac{\lambda_{\alpha}}{\lambda_{p}} = \frac{1}{2\sqrt{2}} \]