Question

The speed of an electron having a wavelength of \[{10^{ - 10}}m\]is:
(A) \[4.24 \times {10^6}m{s^{ - 1}}\]
(B) $5.25 \times {10^6}m{s^{ - 1}}$
(C) $6.26 \times {10^6}m{s^{ - 1}}$
(D) $7.25 \times {10^6}m{s^{ - 1}}$

Answer 1

Hint : Using the wave-particle dual nature find the momentum of the electron from de-Broglie hypothesis. Momentum of a particle is its mass times velocity. Using this relation we get the electron’s velocity.

Formulae used:
De-Broglie relationship between wavelength and momentum of a particle:
$\lambda = \dfrac{h}{p}$...................(1)
Where,
$\lambda $is the wavelength of the particle,
h is planck's constant, $h = 6.63 \times {10^{ - 34}}kg.{m^2}{s^{ - 1}}$
p is momentum of the particle.

Momentum of a particle is defined as:
$p = m.v$ ………………...(2)
Where,
m is mass of the particle,
v is velocity of the particle.

Complete Step by step answer:
Given: The wavelength of the electron is: $\lambda = {10^{ - 10}}m$

To find: Speed of the electron, v.

Step 1
First, put the relation of eq.(2) into eq.(1) to get an expression for v as:
$
  \lambda = \dfrac{h}{{mv}} \\
  \therefore v = \dfrac{h}{{m\lambda }} \\
 $

Step 2
Now, substitute the values of m, h and $\lambda $to get v as:
\[
  v = \dfrac{{6.63 \times {{10}^{ - 34}}kg.{m^2}{s^{ - 1}}}}{{9.1 \times {{10}^{ - 31}}kg \times {{10}^{ - 10}}m}} \\
  \therefore v \simeq 7.25 \times {10^6}m{s^{ - 1}} \\
 \]

Correct answer: The speed of the electron is (d) $7.25 \times {10^6}m{s^{ - 1}}$.

Note: While using the de-broglie wavelength formula for an electron to find its momentum we use the wave property of the electron as only waves can have wavelength. Again when we are using eq.(2) to find the velocity from the momentum then we are using the particle nature of the electron. So, here we are utilizing the wave particle duality of nature.