Question

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 energy of photon is:

• A. $2$
• B. $4$
• C. $\frac{1}{2}$
• D. $\frac{1}{4}$

Answer 1

The correct option is D $\frac{1}{4}$

$K_{particle}=\frac{1}{2}{mV}^2$ also $\lambda =\frac{h}{mV}$

$\Rightarrow K_{particle}=\frac{1}{2}\left(\frac{h}{\lambda v}\right).v^2=\frac{vh}{2\lambda }⟶\left(1\right)$

$K_{photon}=\frac{h_c}{\lambda }$

$\therefore$ $\frac{K_{particle}}{K_{photon}}=\frac{v}{2c}=\frac{1.5\times {10}^8}{2\times 3\times {10}^2}=\frac{1}{4}$