Question

If alpha particle, proton and electron move with the same momentum, then their respective de Broglie wavelengths $\lambda_{\alpha}\lambda_{p}\lambda_{e},are$related as:

• A. $\lambda_{\alpha} = \lambda_{p} = \lambda_{e}$
• B. $\lambda_{\alpha} < \lambda_{p} < \lambda_{e}$
• C. $\lambda_{\alpha} > \lambda_{p} > \lambda_{e}$
• D. $\lambda_{\alpha} > \lambda_{p} > \lambda_{e}$

Answer 1

Answer: (A)

$\lambda_{\alpha} = \lambda_{p} = \lambda_{e}$

Answer 2

: de Broglie wavelength

$\lambda = \frac{h}{p}$

Where symbols have their usual meaning

}{\therefore\lambda_{\alpha} = \lambda_{p} = \lambda_{e}}$$