Step 1: Understanding the Question:
The question asks for the ratio of the de Broglie wavelength of an alpha particle (\(\lambda_\alpha\)) to that of a proton (\(\lambda_p\)) when both particles are accelerated from rest through the same potential difference, \(V\).
Step 2: Key Formula or Approach:
The de Broglie wavelength (\(\lambda\)) is given by \(\lambda = h/p\), where \(h\) is Planck's constant and \(p\) is the momentum. When a particle with charge \(q\) is accelerated through a potential \(V\), its kinetic energy is \(K = qV\). The momentum can be expressed as \(p = \sqrt{2mK}\). Combining these gives:
\[
\lambda = \frac{h}{\sqrt{2mK}} = \frac{h}{\sqrt{2mqV}}
\]
Since \(h\), \(V\), and 2 are constant for both particles, the wavelength is inversely proportional to the square root of the product of mass and charge: \(\lambda \propto \frac{1}{\sqrt{mq}}\).
Step 3: Detailed Explanation:
Let's set up the ratio:
\[
\frac{\lambda_\alpha}{\lambda_p} = \frac{1/\sqrt{m_\alpha q_\alpha}}{1/\sqrt{m_p q_p}} = \sqrt{\frac{m_p q_p}{m_\alpha q_\alpha}}
\]
Now, we need the properties of a proton and an alpha particle:
Proton (\(p\)): Mass \(m_p\), Charge \(q_p = e\).
Alpha particle (\(\alpha\), a Helium nucleus \(^4_2\text{He}\)): Mass \(m_\alpha \approx 4m_p\), Charge \(q_\alpha = 2e\).
Substitute these values into the ratio:
\[
\frac{\lambda_\alpha}{\lambda_p} = \sqrt{\frac{m_p e}{(4m_p)(2e)}} = \sqrt{\frac{1}{8}} = \frac{1}{2\sqrt{2}}
\]
So, the calculated ratio is \(\lambda_\alpha : \lambda_p = 1 : 2\sqrt{2}\).
Step 4: Final Answer:
Thus approximately,
\[ \lambda_\alpha : \lambda_p = 1:2\]