Question

Find the ratio of de-Broglie wavelength of a proton and a 𝝰 - particle, when accelerated through a potential difference of 2V and 4V respectively.

• A. 4:1
• B. 2:1
• C. 1:8
• D. 16:1

Answer 1

The Correct Option is A

To find the ratio of the de-Broglie wavelengths of a proton and an alpha (𝝰) particle when they are accelerated through different potential differences, we will use the de-Broglie wavelength formula:

\(\lambda = \frac{h}{\sqrt{2mEk}}\)

In this formula, \(\lambda\) is the de-Broglie wavelength, \(h\) is Planck's constant, \(m\) is the mass of the particle, and \(Ek\) is the kinetic energy acquired due to acceleration by a potential difference \(V\), which is given by:

\(Ek = qV\)

Step-by-step Calculation:

1. For the proton:
   • Charge \(q = e\) (charge of the proton),
   • Mass \(m_p\),
   • Potential difference \(V = 2V\).
2. For the alpha particle:
   • It has twice the charge of a proton and four times the mass, so \(q = 2e\),
   • Mass \(m_{\alpha} = 4m_p\),
   • Potential difference \(V = 4V\).
3. Now, calculate the ratio of the de-Broglie wavelengths:

\(\frac{\lambda_p}{\lambda_{\alpha}} = \frac{\frac{h}{\sqrt{4m_peV}}}{\frac{h}{\sqrt{16m_peV}}} = \frac{\sqrt{16m_peV}}{\sqrt{4m_peV}} = \frac{\sqrt{16}}{\sqrt{4}} = \frac{4}{2} = 2\)

Thus, the ratio of the de-Broglie wavelengths of the proton and the alpha particle is 2:1.

Therefore, the correct answer from the provided options was misidentified and should be:

2:1