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 $\frac{E_1}{E_2}=\frac{\frac{9}{32}m{c}^2}{mvc}=\frac{\frac{9}{32}m{c}^2}{m\times \frac{3}{4}c\times c}$ $=\frac{9}{32}\times \frac{4}{3}=\frac{3}{8}$ $\frac{E_1}{E_2}=\frac{3}{8}$