Question

An electron, a neutron and an alpha particle have same kinetic energy and their de-Broglie wavelengths are $\lambda_e, \lambda_n$ and $\lambda_\alpha$ respectively. Which statement is correct about their de-Broglie wavelengths?

• A. $\lambda_e > \lambda_n > \lambda_\alpha$
• B. $\lambda_e < \lambda_n > \lambda_\alpha$
• C. $\lambda_e < \lambda_n < \lambda_\alpha$
• D. $\lambda_e > \lambda_n < \lambda_\alpha$
• E. $\lambda_e = \lambda_n < \lambda_\alpha$

Hint:

For same KE, wavelength inversely depends on $\sqrt{m}$.

Answer 1

The Correct Option is A

Concept:
\[ \lambda = \frac{h}{\sqrt{2mK}} \Rightarrow \lambda \propto \frac{1}{\sqrt{m}} \]

Step 1: Mass comparison.
\[ m_e < m_n < m_\alpha \]

Step 2: Relation.
Smaller mass $\Rightarrow$ larger wavelength. \[ \lambda_e > \lambda_n > \lambda_\alpha \]