Question

Find the ratio of de-Broglie wavelengths of deuteron having energy E and \(\alpha\)-particle having energy 2E :

Hint:

Remember that an \(\alpha\)-particle is a Helium nucleus (\(4m_p\)) and a deuteron is an isotope of Hydrogen (\(2m_p\)).
The ratio depends on \(\frac{1}{\sqrt{mK}}\), so \(\frac{\lambda_1}{\lambda_2} = \sqrt{\frac{m_2 K_2}{m_1 K_1}}\).

Answer 1

Correct Answer: 2

Step 1: Understanding the Concept:
The de-Broglie wavelength of a particle is related to its kinetic energy and mass.
As the particle's mass or energy increases, its wavelength decreases.
Step 2: Key Formula or Approach:
The de-Broglie wavelength (\(\lambda\)) is given by:
\[ \lambda = \frac{h}{p} = \frac{h}{\sqrt{2mK}} \]
Where \(h\) is Planck's constant, \(m\) is the mass, and \(K\) is the kinetic energy.
Step 3: Detailed Explanation:
Let \(m_p\) be the mass of a proton.
For a deuteron (\(d\)): Mass \(m_d = 2m_p\) and Energy \(K_d = E\).
\[ \lambda_d = \frac{h}{\sqrt{2(2m_p)E}} = \frac{h}{\sqrt{4m_p E}} \]
For an \(\alpha\)-particle (\(\alpha\)): Mass \(m_{\alpha} = 4m_p\) and Energy \(K_{\alpha} = 2E\).
\[ \lambda_{\alpha} = \frac{h}{\sqrt{2(4m_p)(2E)}} = \frac{h}{\sqrt{16m_p E}} \]
Now, find the ratio:
\[ \frac{\lambda_d}{\lambda_{\alpha}} = \frac{\frac{h}{\sqrt{4m_p E}}}{\frac{h}{\sqrt{16m_p E}}} \]
\[ \frac{\lambda_d}{\lambda_{\alpha}} = \sqrt{\frac{16m_p E}{4m_p E}} = \sqrt{4} = 2 \]
Step 4: Final Answer:
The ratio of the de-Broglie wavelengths of the deuteron and the \(\alpha\)-particle is 2.