A positron and a proton are accelerated by the same accelerating potential. Then the ratio of the associated wavelengths of the positron and the proton will be [M=Mass of proton, m=Mass of positron]

Options

A. `(M)/(m)`
B. `sqrt((M)/(m))`
C. `(m)/(M)`
D. `sqrt((m)/(M))`

Answer

To solve the problem of finding the ratio of the associated wavelengths of a positron and a proton when both are accelerated by the same potential, we can follow these steps:

### Step 1: Write the formula for the de Broglie wavelength
The de Broglie wavelength (λ) associated with a charged particle can be expressed as:
\[ \lambda = \frac{h}{\sqrt{2 m q V}} \]
where:
- \( h \) = Planck's constant
- \( m \) = mass of the particle
- \( q \) = charge of the particle
- \( V \) = accelerating potential

### Step 2: Calculate the wavelength for the positron
For the positron, we have:
- Mass = \( m \) (mass of the positron)
- Charge = \( e \) (charge of the positron)
- Potential = \( V \)

Thus, the wavelength of the positron (\( \lambda_{e^+} \)) is:
\[ \lambda_{e^+} = \frac{h}{\sqrt{2 m e V}} \]

### Step 3: Calculate the wavelength for the proton
For the proton, we have:
- Mass = \( M \) (mass of the proton)
- Charge = \( e \) (charge of the proton)
- Potential = \( V \)

Thus, the wavelength of the proton (\( \lambda_{p} \)) is:
\[ \lambda_{p} = \frac{h}{\sqrt{2 M e V}} \]

### Step 4: Find the ratio of the wavelengths
Now, we need to find the ratio of the wavelengths of the positron to the proton:
\[ \frac{\lambda_{e^+}}{\lambda_{p}} = \frac{\frac{h}{\sqrt{2 m e V}}}{\frac{h}{\sqrt{2 M e V}}} \]

### Step 5: Simplify the ratio
The \( h \), \( 2 \), \( e \), and \( V \) terms cancel out:
\[ \frac{\lambda_{e^+}}{\lambda_{p}} = \frac{\sqrt{2 M e V}}{\sqrt{2 m e V}} = \sqrt{\frac{M}{m}} \]

### Final Result
Thus, the ratio of the associated wavelengths of the positron and the proton is:
\[ \frac{\lambda_{e^+}}{\lambda_{p}} = \sqrt{\frac{M}{m}} \]