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:
\[ \lambda = \frac{h}{p} \]
where \( h \) is Planck's constant and \( p \) is the momentum of the particle.

2. **Relate momentum to kinetic energy**:
The momentum \( p \) can be expressed in terms of kinetic energy (K.E.):
\[ p = \sqrt{2mK.E.} \]
For a charged particle accelerated through a potential difference \( V \), the kinetic energy gained is:
\[ K.E. = qV \]
Thus, we can rewrite momentum as:
\[ p = \sqrt{2mqV} \]

3. **Substitute into the de Broglie wavelength formula**:
Substituting the expression for momentum into the de Broglie wavelength formula gives:
\[ \lambda = \frac{h}{\sqrt{2mqV}} \]

4. **Set up the equation for both particles**:
For the proton and the alpha particle, we have:
\[ \lambda_{proton} = \lambda_{alpha} \]
Therefore:
\[ \frac{h}{\sqrt{2m_{p}q_{p}V_{p}}} = \frac{h}{\sqrt{2m_{\alpha}q_{\alpha}V_{\alpha}}} \]
Here, \( V_{p} = 29.6 \, V \) (the potential difference for the proton), \( m_{p} \) and \( q_{p} \) are the mass and charge of the proton, and \( m_{\alpha} \) and \( q_{\alpha} \) are the mass and charge of the alpha particle.

5. **Cancel \( h \) and rearrange**:
By canceling \( h \) and rearranging, we get:
\[ \sqrt{2m_{p}q_{p}V_{p}} = \sqrt{2m_{\alpha}q_{\alpha}V_{\alpha}} \]
Squaring both sides results in:
\[ 2m_{p}q_{p}V_{p} = 2m_{\alpha}q_{\alpha}V_{\alpha} \]

6. **Solve for \( V_{\alpha} \)**:
Rearranging gives:
\[ V_{\alpha} = \frac{m_{p}q_{p}V_{p}}{m_{\alpha}q_{\alpha}} \]

7. **Substitute known values**:
- Mass of proton \( m_{p} = 1 \, \text{amu} \)
- Mass of alpha particle \( m_{\alpha} = 4 \, \text{amu} \)
- Charge of proton \( q_{p} = 1e \)
- Charge of alpha particle \( q_{\alpha} = 2e \)
- Given \( V_{p} = 29.6 \, V \)

Substituting these values:
\[ V_{\alpha} = \frac{(1)(1)(29.6)}{(4)(2)} = \frac{29.6}{8} = 3.7 \, V \]

### Final Answer:
The potential difference required to accelerate the alpha particle to have the same wavelength as the proton is **3.7 V**.