A proton, when accelerated through a potential differnece of V = 29.6 V, has a wavelength `lambda` associated with it. An `alpha-` particle, in order to have the same `lambda`, must be accelerated through a potential difference of how many volts?

Answer

To solve the problem, we need to find the potential difference required to accelerate an alpha particle so that it has the same de Broglie wavelength as a proton accelerated through a potential difference of 29.6 V.

### Step-by-step Solution:

1. **Understand the de Broglie Wavelength Formula**:
The de Broglie wavelength (λ) of a particle is given by the formula:
\[ \lambda = \frac{h}{mv} \]
where \( h \) is Planck's constant, \( m \) is the mass of the particle, and \( v \) is its velocity.

2. **Relate Velocity to Potential Difference**:
When a charged particle is accelerated through a potential difference \( V \), it gains kinetic energy equal to the work done on it by the electric field:
\[ K.E. = qV \]
where \( q \) is the charge of the particle. For a particle of mass \( m \) and velocity \( v \), the kinetic energy can also be expressed as:
\[ K.E. = \frac{1}{2}mv^2 \]
Equating the two expressions for kinetic energy gives:
\[ qV = \frac{1}{2}mv^2 \]
Rearranging this, we find:
\[ v = \sqrt{\frac{2qV}{m}} \]

3. **Substituting for Wavelength**:
Substituting \( v \) into the de Broglie wavelength formula gives:
\[ \lambda = \frac{h}{m\sqrt{\frac{2qV}{m}}} = \frac{h}{\sqrt{2mqV}} \]

4. **Apply to Proton and Alpha Particle**:
- For the proton:
\[ \lambda_p = \frac{h}{\sqrt{2m_p q_p V_p}} \]
where \( V_p = 29.6 \, V \).
- For the alpha particle:
\[ \lambda_{\alpha} = \frac{h}{\sqrt{2m_{\alpha} q_{\alpha} V_{\alpha}}} \]

5. **Equate the Wavelengths**:
Since we want the wavelengths to be equal:
\[ \frac{h}{\sqrt{2m_p q_p V_p}} = \frac{h}{\sqrt{2m_{\alpha} q_{\alpha} V_{\alpha}}} \]
Canceling \( h \) and squaring both sides gives:
\[ \frac{1}{2m_p q_p V_p} = \frac{1}{2m_{\alpha} q_{\alpha} V_{\alpha}} \]

6. **Substituting Known Values**:
- The mass of the alpha particle \( m_{\alpha} \) is approximately \( 4m_p \).
- The charge of the alpha particle \( q_{\alpha} \) is \( 2q_p \).
Substituting these values into the equation gives:
\[ \frac{1}{2m_p q_p V_p} = \frac{1}{2(4m_p)(2q_p)V_{\alpha}} \]

7. **Simplifying the Equation**:
Simplifying the equation:
\[ \frac{1}{2m_p q_p V_p} = \frac{1}{16m_p q_p V_{\alpha}} \]
Cross-multiplying gives:
\[ 16V_p = V_{\alpha} \]

8. **Calculate \( V_{\alpha} \)**:
Substituting \( V_p = 29.6 \, V \):
\[ V_{\alpha} = 16 \times 29.6 = 473.6 \, V \]

### Final Answer:
The potential difference required to accelerate the alpha particle to have the same wavelength as the proton is approximately **473.6 V**.