#### An electron traveling at  $1.5\times{10}^7\ m/s$ has a de Broglie wavelength equal to that of a photon. Calculate the ratio of the photon's to the electron's kinetic energy.Option: 1 40Option: 2 60Option: 3 20Option: 4 50

Explanation:

The 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,

$E_1=\frac{hc}{\lambda}$

Formula used:

The de Broglie wavelengths are given by,

$\lambda=\frac{h}{p}$……….(1)

Where h is Planck's constant

P is the momentum of the particle

The momentum of the particles is given by,

$p=mv$……….(2)

The kinetic energy of a particle is given by,

$E=\frac{1}{2} \ mv^2$…….(3)

Where

m is the mass of the particle.

v is the mass of the particle.

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,

$\lambda=\frac{h}{p}$

Putting the values of (p) in equation(1) we have,

$\lambda\ =\frac{h}{mv}$

We know that the energy of a photon is given by,

$E_1=\frac{hc}{\lambda}$

The kinetic energy of the electron is

$E_2=\frac{1}{2}mv^2$

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

$\frac{E_1}{E_2}\\ = hc \times\frac{ \frac{1}{\lambda }}{ \frac{1}{2}}\ mv2 (\text{ c = speed of light}) \\ = 2\frac{c}{v} \\ = 2\times3\times\frac{{10}^8}{1.5}\times{10}^7 \\ = 40$