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.

Options

A. `(1)/(2)`
B. `(1)/(sqrt2)`
C. `(1)/(2sqrt2)`
D. `2sqrt2`

Answer

To find the ratio of the de-Broglie wavelengths associated with an alpha particle and a proton when both are accelerated through the same potential difference \( V \), 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:
\[ \lambda = \frac{h}{p} \]
where \( h \) is Planck's constant and \( p \) is the momentum of the particle.

### Step 2: Relate momentum to kinetic energy
When a charged particle is accelerated through a potential difference \( V \), it gains kinetic energy equal to the work done on it by the electric field:
\[ KE = qV \]
where \( q \) is the charge of the particle. The momentum \( p \) can be expressed in terms of kinetic energy:
\[ p = \sqrt{2m \cdot KE} = \sqrt{2m \cdot qV} \]
where \( m \) is the mass of the particle.

### Step 3: Write the de-Broglie wavelength for both particles
For the alpha particle:
\[ \lambda_{\alpha} = \frac{h}{\sqrt{2m_{\alpha} \cdot (2e)V}} \]
where \( m_{\alpha} \) is the mass of the alpha particle and \( 2e \) is its charge (since an alpha particle consists of 2 protons and 2 neutrons).

For the proton:
\[ \lambda_{p} = \frac{h}{\sqrt{2m_{p} \cdot eV}} \]
where \( m_{p} \) is the mass of the proton and \( e \) is its charge.

### Step 4: Find the ratio of the wavelengths
Now, we can find the ratio of the de-Broglie wavelengths:
\[ \frac{\lambda_{\alpha}}{\lambda_{p}} = \frac{\frac{h}{\sqrt{2m_{\alpha} \cdot (2e)V}}}{\frac{h}{\sqrt{2m_{p} \cdot eV}}} \]
This simplifies to:
\[ \frac{\lambda_{\alpha}}{\lambda_{p}} = \frac{\sqrt{2m_{p} \cdot eV}}{\sqrt{2m_{\alpha} \cdot (2e)V}} = \frac{\sqrt{m_{p}}}{\sqrt{m_{\alpha}}} \cdot \frac{1}{\sqrt{2}} \]

### Step 5: Substitute the known values
The mass of the alpha particle is approximately \( 4m_{p} \) (since it consists of 2 protons and 2 neutrons), and the charge of the alpha particle is \( 2e \):
\[ \frac{\lambda_{\alpha}}{\lambda_{p}} = \frac{\sqrt{m_{p}}}{\sqrt{4m_{p}}} \cdot \frac{1}{\sqrt{2}} = \frac{\sqrt{m_{p}}}{2\sqrt{m_{p}}} \cdot \frac{1}{\sqrt{2}} = \frac{1}{2\sqrt{2}} \]

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