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

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.