Step 1: Conceptualization:
This problem links the de-Broglie wavelength of a charged particle to its kinetic energy acquired via acceleration through a potential difference. The objective is to determine the potential required to give an alpha particle the same de-Broglie wavelength as a proton accelerated through potential V.
Step 2: Governing Principles:
1. Kinetic Energy (KE) from acceleration: \(KE = qV_{acc}\), where \(q\) is the particle's charge and \(V_{acc}\) is the accelerating potential.
2. De-Broglie Wavelength (\(\lambda\)): \(\lambda = h/p\), where \(h\) is Planck's constant and \(p\) is momentum.
3. Momentum and Kinetic Energy relation: \(p = \sqrt{2m(KE)}\), where \(m\) is the particle's mass.
Combining these yields the de-Broglie wavelength as a function of accelerating potential:\[ \lambda = \frac{h}{\sqrt{2mqV_{acc}}} \]
Step 3: Derivation and Calculation:
Let subscript 'p' denote proton properties and '\(\alpha\)' denote alpha particle properties.
- Proton: mass \(m_p\), charge \(q_p = e\).
- Alpha particle (\(^{4}_{2}\text{He}\)): mass \(m_\alpha \approx 4m_p\), charge \(q_\alpha = 2e\).
Proton's wavelength: Accelerated through potential \(V\).\[ \lambda_p = \frac{h}{\sqrt{2m_p q_p V}} = \frac{h}{\sqrt{2m_p e V}} \]Alpha particle's wavelength: Accelerated through potential \(V_\alpha\).\[ \lambda_\alpha = \frac{h}{\sqrt{2m_\alpha q_\alpha V_\alpha}} = \frac{h}{\sqrt{2(4m_p)(2e)V_\alpha}} = \frac{h}{\sqrt{16m_p e V_\alpha}} \]Equating wavelengths (\(\lambda_\alpha = \lambda_p\)):\[ \frac{h}{\sqrt{16m_p e V_\alpha}} = \frac{h}{\sqrt{2m_p e V}} \]Squaring both sides and canceling \(h\):\[ \frac{1}{16m_p e V_\alpha} = \frac{1}{2m_p e V} \]Canceling \(m_p\) and \(e\):\[ \frac{1}{16V_\alpha} = \frac{1}{2V} \]Solving for \(V_\alpha\):\[ 16V_\alpha = 2V \]\[ V_\alpha = \frac{2V}{16} = \frac{V}{8} \]
Step 4: Conclusion:
The alpha particle requires an accelerating potential of V/8.