Q 12-11-166JEE MainJEE Main 2019 (11 Jan, Shift 1)Medium
If the de Broglie wavelength of an electron is equal to $10^{-3}$ times the wavelength of a photon of frequency $6\times10^{14}$ Hz, then the speed of electron is equal to: (Speed of light $= 3\times10^8$ m/s, Planck's constant $= 6.63\times10^{-34}$ J s, mass of electron $= 9.1\times10^{-31}$ kg)
Answer: (D) $1.45\times10^{6}$ m/s
Photon wavelength: $\lambda_p = \dfrac cf = \dfrac{3\times10^8}{6\times10^{14}} = 5\times10^{-7}$ m.
Electron wavelength: $\lambda_e = 10^{-3}\lambda_p = 5\times10^{-10}$ m.
$$v = \frac{h}{m\lambda_e} = \frac{6.63\times10^{-34}}{9.1\times10^{-31}\times5\times10^{-10}} \approx 1.45\times10^{6}\ \text{m/s}$$
Solution by Sreeraj P, M.Sc Physics