Question

If $E_e,E_{\alpha }$ and $E_p$ represents the kinetic energies of an electron, alpha particle and a proton respectively, each moving with same de Broglie wavelength, then:

• A. $E_e=E_{\alpha }=E_p$
• B. $E_e>E_p>E_{\alpha }$
• C. $E_{\alpha }>E_p>E_e$
• D. $E_e=E_p=E_{\alpha }$

Answer 1

The correct option is B $E_e>E_p>E_{\alpha }$

$P=\sqrt{2m\times K.E.}$ and $\lambda =\frac{h}{P}$, where $P$ is the momentum, $K.E.$ is the kinetic energy and $\lambda$ is the wavelength.

Same de Broglie wavelength implies the same momentum.
The decreasing order of the mass of the particles is as follows:

$\alpha >proton>electron$

$K.E.=\frac{P^2}{2m}$

Hence, $E_e>E_p>E_{\alpha }$.