Step 1: Understanding the Concept:
The de-Broglie wavelength of a charged particle accelerated through a potential difference depends on its mass, its charge, and the accelerating voltage.
Step 2: Key Formula or Approach:
The de-Broglie wavelength is:
\[ \lambda = \frac{h}{\sqrt{2mqV}} \]
Step 3: Detailed Explanation:
Let \(m_{p}\) and \(e\) be the mass and charge of a proton.
The \(\alpha\)-particle has mass \(m_{\alpha} = 4m_{p}\) and charge \(q_{\alpha} = 2e\).
For the proton:
\[ \lambda_{0} = \frac{h}{\sqrt{2m_{p}e(100)}} \]
For the \(\alpha\)-particle:
\[ \lambda_{\alpha} = \frac{h}{\sqrt{2(4m_{p})(2e)(800)}} \]
Take the ratio \(\frac{\lambda_{\alpha}}{\lambda_{0}}\):
\[ \frac{\lambda_{\alpha}}{\lambda_{0}} = \sqrt{\frac{2m_{p}e(100)}{2(4m_{p})(2e)(800)}} \]
\[ \frac{\lambda_{\alpha}}{\lambda_{0}} = \sqrt{\frac{100}{4 \times 2 \times 800}} = \sqrt{\frac{100}{6400}} = \sqrt{\frac{1}{64}} \]
\[ \lambda_{\alpha} = \frac{\lambda_{0}}{8} \]
Step 4: Final Answer:
The de-Broglie wavelength of the \(\alpha\)-particle is \(\lambda_{0}/8\).