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

Concept: The de Broglie wavelength of a particle is given by: \[ \lambda=\frac{h}{\sqrt{2mK}} \] where \(m\) is the mass and \(K\) is the kinetic energy. Thus: \[ \lambda \propto \frac{1}{\sqrt{mK}} \]
Step 1: Determine masses of the particles
Deuteron consists of one proton and one neutron: \[ m_d=2m \]
Alpha particle consists of two protons and two neutrons: \[ m_\alpha=4m \]
Step 2: Write expressions for wavelengths Deuteron: \[ \lambda_d \propto \frac{1}{\sqrt{2m\cdot E}} \] Alpha particle: \[ \lambda_\alpha \propto \frac{1}{\sqrt{4m\cdot2E}}=\frac{1}{\sqrt{8mE}} \]
Step 3: Take the ratio \[ \frac{\lambda_d}{\lambda_\alpha} =\sqrt{\frac{8mE}{2mE}} =\sqrt{4}=2 \]
Step 4: Express in required form \[ n:1=2:1 \Rightarrow n=2 \] Final Answer: \[ \boxed{2} \]