Question

The de-Broglie wavelength of an electron is the same as that of a photon. If velocity of electron is 25% of the velocity of light, then the ratio of K.E. of electron and K.E. of photon will be:

(1) \(\frac{1}{1}\)

(2) \(\frac{1}{8}\)

(3) \(\frac{8}{1}\)

(4) \(\frac{1}{4}\)

Answer 1

The correct option is (2) \(\frac{1}{8}\)

For photon

\(E_p = \frac{hc}{\lambda_p}\Rightarrow \lambda_p = \frac{hc}{E_p}\)

For electron

\(\lambda_e = \frac{h}{m_ev_e} = \frac{hv_e}{2K_e}\)

Given \(v_e = 0.25 c\)

\(\lambda_e = \frac{h \times 0.25c}{2K_e} = \frac{hc}{8K_e}\)

Also \(\lambda_p = \lambda_e\)

\(\frac{hc}{E_p} = \frac{hc}{8K_e}\)

\(\frac{K_e}{E_p} = \frac{1}{8}\)