Question

The ratio of de Broglie wavelength of a deuteron with kinetic energy \(E\) to that of an alpha particle with kinetic energy \(2E\) is \(n:1\). (Assume mass of proton \(=\) mass of neutron.) Find the value of \(n\).

Hint:

For de Broglie wavelength comparisons, focus on the product \(mK\); constants cancel out immediately.

Answer 1

Correct Answer: 2

Step 1: Understanding the Concept:
The de Broglie wavelength of a particle is inversely proportional to its momentum. For a particle with mass \(m\) and kinetic energy \(K\), the wavelength is related to these parameters through the energy-momentum relation.
Step 2: Key Formula or Approach:
De Broglie wavelength: \(\lambda = \frac{h}{p} = \frac{h}{\sqrt{2mK}}\).
Step 3: Detailed Explanation:
Let \(m_p\) be the mass of a proton.
For a deuteron (\(^2_1H\)): Mass \(m_d \approx 2m_p\). Kinetic energy \(K_d = E\). \[ \lambda_d = \frac{h}{\sqrt{2(2m_p)E}} = \frac{h}{\sqrt{4m_pE}} \] For an alpha particle (\(^4_2He\)): Mass \(m_\alpha \approx 4m_p\). Kinetic energy \(K_\alpha = 2E\). \[ \lambda_\alpha = \frac{h}{\sqrt{2(4m_p)(2E)}} = \frac{h}{\sqrt{16m_pE}} \] Taking the ratio: \[ \frac{\lambda_d}{\lambda_\alpha} = \frac{\frac{h}{\sqrt{4m_pE}}}{\frac{h}{\sqrt{16m_pE}}} = \sqrt{\frac{16m_pE}{4m_pE}} = \sqrt{4} = 2 \] The ratio is given as \(n : 1\), so \(n = 2\).
Step 4: Final Answer:
The value of \(n\) is 2.