Question

A proton accelerated through a potential difference of V volts has a de-Broglie wavelength \(\lambda\) associated with it. In order to get the same wavelength associated with an \(\alpha\)-particle, the required accelerating potential is

• A. V/16
• B. V/8
• C. 4V
• D. 8V

Hint:

From the formula \(\lambda = \frac{h}{\sqrt{2mqV_{acc}}}\), we can see that for a constant wavelength \(\lambda\), the product \(mqV_{acc}\) must be constant. So, \(m_p q_p V_p = m_\alpha q_\alpha V_\alpha\). Plugging in \(m_\alpha = 4m_p\) and \(q_\alpha = 2q_p\), we get \(m_p q_p V = (4m_p)(2q_p)V_\alpha\), which simplifies to \(V = 8V_\alpha\) or \(V_\alpha = V/8\). This ratio method is very fast for comparison problems.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
This problem connects the de-Broglie wavelength of a charged particle with the kinetic energy it gains when accelerated through a potential difference. We need to find the potential required for an alpha particle to have the same wavelength as a proton accelerated through potential V.

Step 2: Key Formula or Approach:
1. The kinetic energy (KE) gained by a particle with charge \(q\) accelerated through a potential difference \(V_{acc}\) is \(KE = qV_{acc}\). 2. The de-Broglie wavelength (\(\lambda\)) is related to momentum (\(p\)) by \(\lambda = h/p\). 3. Momentum is related to kinetic energy by \(p = \sqrt{2m(KE)}\). Combining these, we get an expression for the de-Broglie wavelength in terms of the accelerating potential: \[ \lambda = \frac{h}{\sqrt{2m(KE)}} = \frac{h}{\sqrt{2mqV_{acc}}} \]

Step 3: Detailed Explanation:
Let's denote the properties of the proton with subscript 'p' and the alpha particle with subscript '\(\alpha\)'. - Proton: mass \(m_p\), charge \(q_p = e\). - Alpha particle (\(^{4}_{2}\text{He}\)): mass \(m_\alpha \approx 4m_p\), charge \(q_\alpha = 2e\). For the proton: It is accelerated through potential \(V\). Its wavelength is: \[ \lambda_p = \frac{h}{\sqrt{2m_p q_p V}} = \frac{h}{\sqrt{2m_p e V}} \] For the alpha particle: Let it be accelerated through a potential \(V_\alpha\). Its wavelength is: \[ \lambda_\alpha = \frac{h}{\sqrt{2m_\alpha q_\alpha V_\alpha}} = \frac{h}{\sqrt{2(4m_p)(2e)V_\alpha}} = \frac{h}{\sqrt{16m_p e V_\alpha}} \] Equating the wavelengths: We are given that we want the same wavelength, so \(\lambda_\alpha = \lambda_p\). \[ \frac{h}{\sqrt{16m_p e V_\alpha}} = \frac{h}{\sqrt{2m_p e V}} \] Squaring both sides (and canceling \(h\)): \[ \frac{1}{16m_p e V_\alpha} = \frac{1}{2m_p e V} \] The terms \(m_p\) and \(e\) cancel out. \[ \frac{1}{16V_\alpha} = \frac{1}{2V} \] Rearranging to solve for \(V_\alpha\): \[ 16V_\alpha = 2V \] \[ V_\alpha = \frac{2V}{16} = \frac{V}{8} \]

Step 4: Final Answer:
The required accelerating potential for the alpha particle is V/8.