Question

The kinetic energy of an electron is \[4.55\times {{10}^{-25}}\text{J}\]. Calculate the wavelength.
(\[\text{h=6}\text{.6 }\!\!\times\!\!\text{ 1}{{\text{0}}^{\text{-34}}}\text{Jsec}\], mass of electron =\[\text{9}\text{.1 }\!\!\times\!\!\text{ 1}{{\text{0}}^{\text{-31}}}\text{kg}\] )

Answer 1

Hint: The wavelength of an electron can be calculated by using the equation of the kinetic energy and de-Broglie. By using the kinetic energy, calculate the velocity of the electron and later by substituting that velocity in the de-Broglie equation, we can calculate the wavelength.

Complete step by step solution:
By knowing the kinetic energy and de-Broglie equation it is easy to determine the wavelength of the electron. Before going to that let’s first understand about kinetic energy and de-Broglie equation.
What is kinetic energy? Kinetic energy is defined as the energy possessed by a body by virtue of its motion. Kinetic energy is the work required to accelerate a body of a given mass from rest into motion. It is given
 \[\text{K}\text{.E =}\frac{\text{1}}{\text{2}}\text{m}{{\text{v}}^{\text{2}}}\]
The De-Broglie equation said that matter can act as both matter and particle. de-Broglie equation is \[\lambda =\frac{h}{mv}\]
Where h is the Planck’s constant.
m is the mass of the electron
v is the velocity
\[\lambda \]is the wavelength
Let’s now come to the problem,
In the problem kinetic energy, mass of electron and Planck’s constant are given.
\[K.E=4.55\times {{10}^{-25}}\text{J}\]
 \[\text{m=9}\text{.1 }\!\!\times\!\!\text{ 1}{{\text{0}}^{\text{-31}}}\text{kg}\]
\[\text{h=6}\text{.6 }\!\!\times\!\!\text{ 1}{{\text{0}}^{\text{-34}}}\text{Jsec}\]
First calculate the velocity using kinetic energy
\[\text{K}\text{.E =}\frac{\text{1}}{\text{2}}\text{m}{{\text{v}}^{\text{2}}}\]
\[4.55\times {{10}^{-25}}\text{=}\frac{1}{2}\times 9.1\times {{10}^{-31}}\times {{v}^{2}}\]
\[{{v}^{2}}\text{=}\frac{2\times 4.55\times {{10}^{-25}}}{9.1\times {{10}^{-31}}}\]
\[{{v}^{2}}=1\times {{10}^{6}}m/s\]
\[v=1\times {{10}^{3}}m/s\]
Velocity of the electron is \[v=1\times {{10}^{3}}m/s\].
Substituting the velocity, we found using the kinetic energy equation in the de-Broglie equation.
de-Broglie equation is \[\lambda =\frac{h}{mv}\]
\[\lambda =\frac{6.6\times {{10}^{-25}}}{9.1\times {{10}^{-31}}\times 1\times {{10}^{3}}}\]
\[\lambda =7.28\times {{10}^{-7}}m\]

Therefore, the wavelength of an electron is\[\lambda =7.28\times {{10}^{-7}}m\].

Additional information:
Comparison between kinetic energy and potential energy

Kinetic energy	Potential energy
Energy possessed by a body due to its state of motion.	Energy possessed by a body due to its change in position.
It can be easily transferred from one moving body to another.	It cannot be easily transferred.
\[\text{K}\text{.E =}\frac{\text{1}}{\text{2}}\text{m}{{\text{v}}^{\text{2}}}\]	\[\text{P}\text{.E=mgh}\]

Note: The de-Broglie equation states that matter (i.e. electron) have dual character. It can act as a wave as well as particles. This equation tells us that a beam of electrons can be diffracted just like a beam of light. This equation helps us to understand the idea of matter having wavelength.