Q.23 The de-Broglie wavelengths associated with an electron and proton are same. Calculate the ratio of their momentum and kinetic energies.
Best answer
Answer: \(p_e : p_p = 1 : 1\); \(K_e : K_p \approx 1835 : 1\).
\(\lambda = \dfrac{h}{p}\). Since \(\lambda_e = \lambda_p\), \(p_e = p_p\), i.e. the ratio of momenta is 1 : 1.
Kinetic energy \(K = \dfrac{p^2}{2m}\). With equal momenta,
\[\frac{K_e}{K_p} = \frac{m_p}{m_e} = \frac{1.672\times10^{-27}}{9.11\times10^{-31}} = 1835\]So the electron has about 1835 times the kinetic energy of the proton.
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