Question

The de-Broglie wavelength of a particle moving with a velocity $ 2.25\times {{10}^{8}}m/s $ is equal to the wavelength of a photon. The ratio of kinetic energy of the particle to the energy of the photon is: (Velocity of light is $ 3\times {{10}^{8}}m/s $ )

• A. $ \frac{1}{8} $
• B. $ \frac{3}{8} $
• C. $ \frac{5}{8} $
• D. $ \frac{7}{8} $

Answer 1

Answer: (B)

Velocity of particle $ v=2.25\times {{10}^{8}} $ $ =\frac{3}{4}\times 3\times {{10}^{8}}=\frac{3}{4}c $ $ (\because \,c=3\times {{10}^{8}}\,m/s) $ de-Broglie wavelength of the particle is given by $ {{\lambda }_{1}}=\frac{h}{mv} $ ?(i) Energy of photon is given by as $ {{E}_{2}}=hv=\frac{hc}{{{\lambda }_{2}}} $ $ {{\lambda }_{2}}=\frac{hc}{{{E}_{2}}} $ ?(ii) but $ {{\lambda }_{1}}={{\lambda }_{2}} $ (given) $ \frac{h}{mv}=\frac{hc}{{{E}_{2}}} $ or $ {{E}_{2}}=mvc $ ?(iii) Now, kinetic energy of particle is $ {{E}_{1}}=\frac{1}{2}m{{v}^{2}} $ $ =\frac{1}{2}m{{\left( \frac{3}{4}c \right)}^{2}} $ $ {{E}_{1}}=\frac{9}{32}m{{c}^{2}} $ ?(iv) From Eqs. (iii) and (iv), we get $ \frac{{{E}_{1}}}{{{E}_{2}}}=\frac{\frac{9}{32}m{{c}^{2}}}{mvc}=\frac{\frac{9}{32}m{{c}^{2}}}{m\times \frac{3}{4}c\times c} $ $ =\frac{9}{32}\times \frac{4}{3}=\frac{3}{8} $ $ \frac{{{E}_{1}}}{{{E}_{2}}}=\frac{3}{8} $