Question

The de Broglie wavelengths associated with an electron and a proton is equal. Prove that the kinetic energy of the electron is greater than that of the proton.

Answer 1

de Brogie wavelength λ = \(\frac{h}{\sqrt {2mK}}\)

We get K = \(\frac{h^2}{{2mλ^2}}\)

K ∝ \(\frac{1}{m}\)

We know that m_p > m_e

_{\(\frac{K_e}{K_p} = \frac {m_p}{m_e} > 1\)}

K_e > K_p