The ratio of de-Broglie wavelength associated with a proton and a `mu`-particle accelerated through the same potential difference 'V' is

Options

A. `1:2`
B. `1:1`
C. `1 : 2√2`
D. `1:4`

Answer

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

### Step 1: Understand the Formula for de-Broglie Wavelength
The de-Broglie wavelength \( \lambda \) of a particle is given by the formula:
\[ \lambda = \frac{h}{\sqrt{2qV m}} \]
where:
- \( h \) is Planck's constant,
- \( q \) is the charge of the particle,
- \( V \) is the potential difference,
- \( m \) is the mass of the particle.

### Step 2: Write the Wavelengths for Proton and Muon
For the proton, the de-Broglie wavelength \( \lambda_p \) is:
\[ \lambda_p = \frac{h}{\sqrt{2q_p V m_p}} \]
For the muon, the de-Broglie wavelength \( \lambda_\mu \) is:
\[ \lambda_\mu = \frac{h}{\sqrt{2q_\mu V m_\mu}} \]

### Step 3: Find the Ratio of the Wavelengths
Now, we can find the ratio of the wavelengths:
\[ \frac{\lambda_p}{\lambda_\mu} = \frac{\frac{h}{\sqrt{2q_p V m_p}}}{\frac{h}{\sqrt{2q_\mu V m_\mu}}} \]
This simplifies to:
\[ \frac{\lambda_p}{\lambda_\mu} = \frac{\sqrt{2q_\mu V m_\mu}}{\sqrt{2q_p V m_p}} \]
The constants \( 2V \) cancel out:
\[ \frac{\lambda_p}{\lambda_\mu} = \sqrt{\frac{q_\mu m_\mu}{q_p m_p}} \]

### Step 4: Substitute Values for Charge and Mass
The charge of the muon \( q_\mu \) is equal to the charge of the electron, which is \( 1.6 \times 10^{-19} \) C, and the charge of the proton \( q_p \) is also \( 1.6 \times 10^{-19} \) C. Thus, we have:
\[ \frac{q_\mu}{q_p} = 1 \]

The mass of the muon \( m_\mu \) is \( 1.88 \times 10^{-28} \) kg, and the mass of the proton \( m_p \) is \( 1.6 \times 10^{-27} \) kg. Therefore:
\[ \frac{m_\mu}{m_p} = \frac{1.88 \times 10^{-28}}{1.6 \times 10^{-27}} = 0.1175 \]

### Step 5: Calculate the Final Ratio
Now substituting these values into the ratio:
\[ \frac{\lambda_p}{\lambda_\mu} = \sqrt{\frac{1 \times 1.88 \times 10^{-28}}{1 \times 1.6 \times 10^{-27}}} = \sqrt{0.1175} \approx 0.343 \]

### Step 6: Finalize the Ratio
Thus, the ratio of the de-Broglie wavelengths is approximately:
\[ \frac{\lambda_p}{\lambda_\mu} \approx 0.343 \]
This can be expressed in terms of a fraction, which is approximately \( \frac{1}{2.9} \).