Question

Find the de Broglie wavelength of an electron with a speed of 0.78c. Take relativistic effects into account.

Answer 1

Answer:

de Broglie wavelength of an electron with speed 0.78 c taking relativistic effects into account is given as:

λ = 1.943 * 10^(-12) m

Explanation:

Given:

v = 0.78 c

we know:

c = speed of light = 3 * 10^8 m/s

mass of electron = m = 9.1 × 10-31 kg

de Broglie wavelength:

In 1924 a French physicist Louis de Broglie assumed that for particles the same relations are valid as for the photon:

(Dual-nature of a particle)

Let the wavelength be = λ

According to de Broglie:

λ = h/p = h/mv

where h is planck's constant = 6.626176 x 10^-34 Js

and p is momentum.

Taking relativistic effects into account, we know that the momentum of the particle changes by a factor 'γ'.

At low speed, γ is almost 1. However, at very high velocity (comparable to light), it has a great effect on momentum.

γ = $\frac{1}{\sqrt{1-\frac{v^{2} }{c^{2} } } }$

γ = 1.6

Now at 0.78 c, considering relativistic effects, we know:

λ = h/γp = h/γ*mv

= (6.62 x 10^(-34))/(1.6*0.78*3*10^(8)*9.1 × 10-31

λ = 1.943 * 10^(-12) m