Solveeit Logo
Login

Question

Question: What is the wavelength of an electron with a mass of \(9\times {{10}^{-31}}kg\) and a speed of \(2.5...

What is the wavelength of an electron with a mass of 9×10−31kg9\times {{10}^{-31}}kg and a speed of 2.5×106ms−12.5\times {{10}^{6}}m{{s}^{-1}}.

Explanation

Solution

An electron with given mass is made to move at a very high speed. The de Broglie wavelength gives the relation between the wavelength and kinetic energy of a body. The kinetic energy depends on the mass and square of velocity. Substituting given values we can calculate the de Broglie wavelength of the electron. Convert the units as required.
Formulas used:
λ=h2mE\lambda =\dfrac{h}{\sqrt{2mE}}

Complete step-by-step solution:
Electrons are negatively charged particles which revolve around the nucleus in an atom. Electrons give out matter waves due to its wave nature and the wavelength of its matter waves is equal to the De-Broglie wavelength. Given, an electron has mass 9×10−31kg9\times {{10}^{-31}}kg and speed 2.5×106ms−12.5\times {{10}^{6}}m{{s}^{-1}}.
The relation between wavelength of electron and kinetic energy is given by De-Broglie's equation to calculate wavelength. Therefore,
λ=h2mE\lambda =\dfrac{h}{\sqrt{2mE}} - (1)
Here, λ\lambda is the wavelength
hh is the Planck’s constant
mm is the mass
EE is the kinetic energy
We know that,
E=12mv2E=\dfrac{1}{2}m{{v}^{2}}
Here, vv is the velocity
To calculate the kinetic energy of electron, w substitute given values in the above equation to get,
E=12×9×10−31×kg×(2.5×106ms−1)2 ⇒E=2.84×10−18J \begin{aligned} & E=\dfrac{1}{2}\times 9\times {{10}^{-31}}\times kg\times {{(2.5\times {{10}^{6}}m{{s}^{-1}})}^{2}} \\\ & \Rightarrow E=2.84\times {{10}^{-18}}J \\\ \end{aligned}
Therefore, the kinetic energy of the electron is 2.84×10−18J2.84\times {{10}^{-18}}J.
Now, substituting given values in eq (1), we get,
λ=6.626×10−342×9×10−28kg×2.84×10−18 ⇒λ=9.2×10−12m \begin{aligned} & \lambda =\dfrac{6.626\times {{10}^{-34}}}{\sqrt{2\times 9\times {{10}^{-28}}kg\times 2.84\times {{10}^{-18}}}} \\\ & \Rightarrow \lambda =9.2\times {{10}^{-12}}m \\\ \end{aligned}
Therefore, the wavelength of the electron is 9.2×10−12m9.2\times {{10}^{-12}}m.

Note: For photons the wavelength is equal to Planck’s constant divided by its momentum. The De-Broglie wavelength is used to represent the wave nature of matter. All matter shows waves as well as particle-like behavior. Wavelengths associated with objects with significant sizes have small wavelengths which are irrelevant in our day to day lives.