Question

The de-Broglie wavelength of an electron moving with a velocity of $1.5\times {10}^{8}m/s$ is equal to that of a photon. The ratio of kinetic energy of the electron to that of the photon $(c=3\times {10}^{8}m/s)$:

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

Answer 1

The correct option is C $\frac{1}{2}$

Since the de-Broglie wavelength depends only on the momentum of particle, both electron and photon must have same momentum. Also, kinetic energy expression can be written as,

$K=\frac{1}{2}m{v}^2=\frac{1}{2}pv$

Hence,

$\frac{E_{electron}}{E_{photon}}=\frac{v_{electron}}{v_{photon}}=\frac{1.5\times {10}^8}{3\times {10}^8}=\frac{1}{2}$