Question

The de Broglie wavelength of an electron with a velocity of $1.5\times {{10}^{8}}$ is equal to that of a photon. Find the ratio of kinetic energy of the photon to that of the electron

Answer 1

Hint: Kinetic energy of the photon is equal to the natural energy. Use de Broglie’s hypothesis to determine the wavelength of the photon. Find the energy of the photon using the equation,

${{E}_{1}}=\dfrac{hc}{\lambda }$

Formula Used:
The de Broglie wavelength is given by,

$\lambda =\dfrac{h}{p}$………………….(1)

Where, h is the Planck’s constant
p is the momentum of the particle.
Momentum of the particle is given by,

$p=mv$……………….(2)

Kinetic energy of a particle is given by,

$E=\dfrac{1}{2}m{{v}^{2}}$…………….(3)

Where,
m is the mass of the particle.
v is the velocity of the particle.

Complete step by step answer:
It is given that the de Broglie wavelength of the electron and photon are the same.
So, we need to find the wavelength of the photon to determine the kinetic energy.
Using Equation (1) we get,

$\lambda =\dfrac{h}{p}$
Putting the value of p in equation (1) we have,
$\lambda =\dfrac{h}{mv}$
We know that, energy of a photon is given by,

${{E}_{1}}=\dfrac{hc}{\lambda }$

Kinetic energy of the electron is,

${{E}_{2}}=\dfrac{1}{2}m{{v}^{2}}$

So, the ratio of the kinetic energy of the photon to that of the electron is given by,

$\dfrac{{{E}_{1}}}{{{E}_{2}}}$
= $\dfrac{hc\times (\dfrac{1}{\lambda })}{\dfrac{1}{2}m{{v}^{2}}}$
= $\dfrac{hc\times (\dfrac{mv}{h})}{\dfrac{1}{2}m{{v}^{2}}}$
= $\dfrac{2c}{v}$
=$\dfrac{2\times (3\times {{10}^{8}})}{(1.5\times {{10}^{8}})}$
= 4
Hence, the answer is 4.

Note:
According to de Broglie Hypothesis, the wave nature of small and high-speed particles. According to the wave theory, the wave takes the energy from one place to another with group velocity. Matter waves also follow this condition.