Question

A proton and an electron are associated with same de-Broglie wavelength. The ratio of their kinetic energies is:

(Assume h \(=6.63 \times 10^{-34} \mathrm{~J} \mathrm{~s}, \mathrm{~m}_{\mathrm{e}}=9.0 \times 10^{-31} \mathrm{~kg}\) and \(\mathrm{m}_{\mathrm{p}}=1836\) times \(\mathrm{m}_{\mathrm{e}}\)) 

(1) \(1: 1836\)

(2) \(1: \frac{1}{1836}\)

(3) \(1: \frac{1}{\sqrt{1836}}\)

(4) \(1: \sqrt{1836}\)

Answer 1

Correct option is :  (1) \(1: 1836\)  

\(\lambda\) is same for both

\(\mathrm{P}=\frac{\mathrm{h}}{\lambda}\) same for both

\(\mathrm{P}=\sqrt{2 \mathrm{mK}}\)

Hence,

\(\mathrm{K} \propto \frac{1}{\mathrm{~m}}\)

\(\Rightarrow \frac{\mathrm{KE}_{\mathrm{p}}}{\mathrm{KE}_{\mathrm{e}}}=\frac{\mathrm{m}_{\mathrm{e}}}{\mathrm{m}_{\mathrm{p}}}=\frac{1}{1836}\)