• # question_answer The de Broglie wavelength of an electron moving with a velocity $1.5\times {{10}^{8}}m\,{{s}^{-1}}$is equal to that of a photon. The ratio of the kinetic energy of the electron to that of the photon is A)  2                                B)  4C)  $\frac{1}{2}$                                   D)  $\frac{1}{4}$

The speed of photon$=c=3\times {{10}^{8}}\,m/s,$ Wavelength of photon is v. Wavelength of electron $=\frac{h}{mv}.$ Now, $\frac{\text{KE}\,\text{of}\,\text{electron}}{\text{KE}\,\text{of}\,\text{photon}}=\frac{\frac{1}{2}m{{v}^{2}}}{hv}=\frac{1}{2}\frac{m{{v}^{2}}}{hc}\lambda =\frac{m{{v}^{2}}h}{2hcmv}$ $=\frac{v}{2c}=\frac{1}{4}\,(\because \,v=c/2)$ Hence, the correction option is [d].