Question

A proton, an electron and an alpha particle have the same energies. Their de-Broglie wavelengths will be compared as: 

(1) \(\lambda_{\mathrm{e}}>\lambda_{\alpha}>\lambda_{\mathrm{p}}\)

(2) \(\lambda_{\alpha}<\lambda_{\mathrm{p}}<\lambda_{\mathrm{e}}\)

(3) \(\lambda_{\mathrm{p}}<\lambda_{\mathrm{e}}<\lambda_{\alpha}\)

(4) \(\lambda_{\mathrm{p}}>\lambda_{\mathrm{e}}>\lambda_{\alpha}\)   

Answer 1

Correct option is : (2) \(\lambda_{\alpha}<\lambda_{\mathrm{p}}<\lambda_{\mathrm{e}}\)  

\(\lambda_{\mathrm{DB}}=\frac{\mathrm{h}}{\mathrm{p}}=\frac{\mathrm{h}}{\sqrt{2 \mathrm{mKE}}}\)

KE is same for all.

\(\Rightarrow \lambda\times\frac{1}{\sqrt{\mathrm{m}}}\)

\(\Rightarrow m_{\mathrm{e}}<m_{\mathrm{p}}<m_{\mathrm{\alpha}}\)

\(\Rightarrow \lambda_{\mathrm{\alpha}}<\lambda_{\mathrm{p}}<\lambda_{\mathrm{e}}\)