A proton when accelerated through a potential difference of V volt has a wavelength `lamda` associated with it. An alpha-particle in order to have the same `lamda` must be accelerated through a potential difference of

Options

A. V volt
B. 4V vold
C. 2 V volt
D. `((V)/(8))` volt

Explanation

To solve the problem of determining the potential difference required to accelerate an alpha particle to have the same wavelength as a proton, we can follow these steps:

### Step 1: Understand the Relationship Between Wavelength and Momentum
The de Broglie wavelength (λ) associated with 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
The momentum \( p \) of a particle can be expressed in terms of its mass \( m \) and velocity \( v \):
\[ p = mv \]
The kinetic energy (KE) of the particle is given by:
\[ KE = \frac{1}{2} mv^2 \]
From this, we can express \( v \) in terms of kinetic energy:
\[ v = \sqrt{\frac{2 \cdot KE}{m}} \]

### Step 3: Substitute Kinetic Energy with Charge and Voltage
The kinetic energy 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. Thus, we can rewrite the velocity as:
\[ v = \sqrt{\frac{2QV}{m}} \]

### Step 4: Substitute into the Wavelength Equation
Substituting the expression for \( v \) into the wavelength formula gives:
\[ \lambda = \frac{h}{mv} = \frac{h}{m \sqrt{\frac{2QV}{m}}} = \frac{h}{\sqrt{2mQV}} \]

### Step 5: Set the Wavelengths Equal for Proton and Alpha Particle
Let \( \lambda_p \) be the wavelength of the proton and \( \lambda_{\alpha} \) be the wavelength of the alpha particle. We have:
\[ \lambda_p = \lambda_{\alpha} \]
This implies:
\[ \frac{h}{\sqrt{2m_pQ_pV}} = \frac{h}{\sqrt{2m_{\alpha}Q_{\alpha}V_{\alpha}}} \]
Cancelling \( h \) and rearranging gives:
\[ \sqrt{2m_pQ_pV} = \sqrt{2m_{\alpha}Q_{\alpha}V_{\alpha}} \]

### Step 6: Square Both Sides
Squaring both sides results in:
\[ 2m_pQ_pV = 2m_{\alpha}Q_{\alpha}V_{\alpha} \]
This simplifies to:
\[ m_pQ_pV = m_{\alpha}Q_{\alpha}V_{\alpha} \]

### Step 7: Solve for \( V_{\alpha} \)
Rearranging for \( V_{\alpha} \) gives:
\[ V_{\alpha} = \frac{m_pQ_p}{m_{\alpha}Q_{\alpha}} V \]

### Step 8: Substitute Values for Masses and Charges
For a proton:
- Mass \( m_p \) = \( 1 \, \text{u} \)
- Charge \( Q_p \) = \( e \)

For an alpha particle:
- Mass \( m_{\alpha} \) = \( 4 \, \text{u} \)
- Charge \( Q_{\alpha} \) = \( 2e \)

Substituting these values:
\[ V_{\alpha} = \frac{(1)(e)}{(4)(2e)} V = \frac{1}{8} V \]

### Final Answer
Thus, the potential difference \( V_{\alpha} \) required for the alpha particle to have the same wavelength as the proton is:
\[ V_{\alpha} = \frac{V}{8} \]