A proton is accelerated through a potential difference of 400 V. to have same de-broglie wavelength, what potential difference must be applied across doubly ionised `._(8)O^(16)` atom.

Options

A. 50 volt
B. 12.5 volt
C. 100 volt
D. none of these

Answer

To solve the problem, we need to determine the potential difference that must be applied across a doubly ionized \( ^{16}_{8}O \) atom so that it has the same de Broglie wavelength as a proton accelerated through a potential difference of 400 V.

### Step-by-Step Solution:

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.

2. **Relate momentum to kinetic energy**:
The kinetic energy \( KE \) gained by a charged particle when accelerated through a potential difference \( V \) is given by:
\[ KE = QV \]
where \( Q \) is the charge of the particle. For a proton, \( Q = e \) (the elementary charge).

3. **Express momentum in terms of kinetic energy**:
The momentum \( p \) of a particle can also be expressed in terms of its kinetic energy:
\[ KE = \frac{p^2}{2m} \implies p = \sqrt{2m \cdot KE} \]
Thus, for a proton accelerated through a potential difference \( V_p = 400 \, V \):
\[ p_p = \sqrt{2m_p \cdot e \cdot 400} \]

4. **Calculate the momentum for the doubly ionized \( ^{16}_{8}O \) atom**:
For the doubly ionized \( ^{16}_{8}O \) atom, the charge \( Q_O = 2e \) (since it is doubly ionized) and its mass \( m_O = 16m_p \) (where \( m_p \) is the mass of the proton). The momentum can be expressed as:
\[ p_O = \sqrt{2m_O \cdot Q_O \cdot V_O} \]
where \( V_O \) is the potential difference we need to find.

5. **Set the momenta equal for the same de Broglie wavelength**:
Since we want the de Broglie wavelengths to be the same, we set \( p_p = p_O \):
\[ \sqrt{2m_p \cdot e \cdot 400} = \sqrt{2 \cdot 16m_p \cdot 2e \cdot V_O} \]

6. **Square both sides to eliminate the square roots**:
\[ 2m_p \cdot e \cdot 400 = 2 \cdot 16m_p \cdot 2e \cdot V_O \]

7. **Cancel common terms**:
Cancel \( 2m_p \) and \( e \) from both sides:
\[ 400 = 32V_O \]

8. **Solve for \( V_O \)**:
\[ V_O = \frac{400}{32} = 12.5 \, V \]

### Final Answer:
The potential difference that must be applied across the doubly ionized \( ^{16}_{8}O \) atom is **12.5 V**.