Question

The de-Broglie wavelength of an electron, $\alpha$-patiicle and a proton all having the same kinetic energy is respectively given as $\lambda_e, \, \lambda_\alpha$ and $\lambda_p$. Then which of the following is not true?

• A. $\lambda_e < \lambda_p$
• B. $\lambda_p < \lambda_\alpha$
• C. $\lambda_e > \lambda_\alpha$
• D. $\lambda_\alpha < \lambda_p > \lambda_e$

Answer 1

Answer: (A)

$\lambda_e < \lambda_p$

Answer 2

$\lambda = \frac{ h }{ m v } \,\,\,... (i)$ or $\lambda \propto \frac{ 1}{ m }$ Therefore, $\lambda_e \propto \frac{ 1}{ m_e }, \, \lambda_\alpha \propto \frac{ 1}{ m_\alpha }$ and $\lambda_p \propto \frac{ 1}{ m_p}$ As we know that $m_e < m_p < m_\alpha$ $\therefore \lambda_e > \lambda_p > \lambda_\alpha$ or $\lambda_e > \lambda_\alpha$ or $\lambda_p > \lambda_\alpha$ or $\lambda_e > \lambda_p$