`Be^(3+)` and a proton are accelerated by the same potential, their de`-` Broglie wavelengths have the ratio ( assume mass of proton `=` mass of neutron `) :`

Options

A. `1:2`
B. `1:4`
C. `1:1`
D. `1:3sqrt(3)`

Explanation

To find the ratio of the de Broglie wavelengths of \( \text{Be}^{3+} \) and a proton when both are accelerated by the same potential, 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:
\[ \lambda = \frac{h}{\sqrt{2m q V}} \]
where:
- \( h \) is Planck's constant,
- \( m \) is the mass of the particle,
- \( q \) is the charge of the particle,
- \( V \) is the accelerating potential.

### Step 2: Define the parameters for both particles
1. **For the proton**:
- Mass: \( m_p \)
- Charge: \( q_p = +e \) (where \( e \) is the elementary charge)

2. **For \( \text{Be}^{3+} \)**:
- Mass: \( m_{Be} = 9m_p \) (since the atomic mass of beryllium is approximately 9)
- Charge: \( q_{Be} = +3e \)

### Step 3: Write the de Broglie wavelength for both particles
1. **For the proton**:
\[ \lambda_p = \frac{h}{\sqrt{2m_p q_p V}} = \frac{h}{\sqrt{2m_p e V}} \]

2. **For \( \text{Be}^{3+} \)**:
\[ \lambda_{Be} = \frac{h}{\sqrt{2m_{Be} q_{Be} V}} = \frac{h}{\sqrt{2(9m_p)(3e)V}} = \frac{h}{\sqrt{54m_p e V}} \]

### Step 4: Find the ratio of the wavelengths
To find the ratio \( \frac{\lambda_{Be}}{\lambda_p} \):
\[ \frac{\lambda_{Be}}{\lambda_p} = \frac{\frac{h}{\sqrt{54m_p e V}}}{\frac{h}{\sqrt{2m_p e V}}} \]
This simplifies to:
\[ \frac{\lambda_{Be}}{\lambda_p} = \frac{\sqrt{2m_p e V}}{\sqrt{54m_p e V}} = \frac{\sqrt{2}}{\sqrt{54}} = \frac{\sqrt{2}}{3\sqrt{6}} = \frac{1}{3\sqrt{27}} = \frac{1}{3\sqrt{3^3}} = \frac{1}{3\sqrt{3} \cdot 3} = \frac{1}{3\sqrt{3}} \]

### Step 5: Finalize the answer
Thus, the ratio of the de Broglie wavelengths is:
\[ \frac{\lambda_{Be}}{\lambda_p} = \frac{1}{3\sqrt{3}} \]