Question

If a proton and an electron have same de Broglie wavelengths, then :

• A. Proton has K.E. than electron
• B. electron has more K.E. than proton
• C. both have same K.E.
• D. both have same velocity.

Answer 1

The correct option is B electron has more K.E. than proton

We know, $K.E=\frac{1}{2}\frac{p^2}{m}$

$p=\sqrt{2mK}$

De - Brogile wavelength,

$\lambda =\frac{h}{p}=\frac{h}{\sqrt{2mK}}$

$p=\frac{h}{\lambda }⟶\left(1\right)$

$K=\frac{h^2}{2m{\lambda }^2}⟶\left(2\right)$

If $\lambda$ is constant, then from equation $\left(1\right)$ $p=$ constant

$m_p{V}_p=m_e{V}_e$

$\frac{V_p}{V_e}=\frac{m_e}{m_p}<1$

$V_p<V_e$

If $\lambda$ is constant then from $\left(2\right)$ $K\alpha \frac{1}{m}$

$\frac{K_p}{K_e}=\frac{m_e}{m_p}<1$

$K_p<K_e$

(B) Electron has more $K.E$ than proton $\Rightarrow$ correct answer.