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Question: Calculate the de-Broglie wavelength of an electron beam acceleration through a potential difference ...

Calculate the de-Broglie wavelength of an electron beam acceleration through a potential difference of 60 V.
A.1.58 A
B.2.58 A
C.3.58 A
D.4.58 A

Explanation

Solution

We will make use of the de-Broglie wave equation. Firstly, we will calculate the energy of the particle. Then by substituting the values of Planck's constant, mass of electron and energy in the de-Broglie wavelength formula we will get the required answer.

Formula used: λ=h2mE\lambda =\dfrac{h}{\sqrt{2mE}}
E=qΔV E=q\Delta V\text{ }

Complete step by step solution:
We know the equation derived by the scientist Louis de-Broglie which describes the wave nature of any particle. Therefore, the wavelength of any moving body/object is derived by a formula:
λ=h2mE ............(1)\lambda =\dfrac{h}{\sqrt{2mE}}\text{ }............\text{(1)}
Where λ\lambda denotes the wavelength of a moving body, h is the Planck's constant, m is the mass of the particle in kg and E is the energy of the particle.
In the question we have given the potential difference 60 V, that is ΔV=60V\Delta V=60V
Now, Energy of the particle is given by:
E=qΔV ..........(2)E=q\Delta V\text{ }..........\text{(2)}
We know that charge on one electron is equal to 1.6×10−19C1.6\times {{10}^{-19}}C, so substituting this in equation(1)
we get,
E=1.6×10−19×60 ⇒E=9.6×10−18J .............(3) \begin{aligned} & E=1.6\times {{10}^{-19}}\times 60 \\\ & \Rightarrow E=9.6\times {{10}^{-18}}J\text{ }.............\text{(3)} \\\ \end{aligned}
And momentum, p=mv=2mE ............(4)\sqrt{2mE}\text{ }............\text{(4)}
We know that mass of an electron is m=9.1×10−31kgm=9.1\times {{10}^{-31}}kg,
E=9.6×10−18JoulesE=9.6\times {{10}^{-18}}Joulesand
h=6.62×10−34 Joule secondh=6.62\times {{10}^{-34}}\text{ Joule second}
Substituting the above values in equation (1) we get,
λ=6.62×10−342×9.1×10−31×9.6×10−18 ⇒λ=1.58×10−10m ∴λ=1.58 Angstrom \begin{aligned} & \lambda =\dfrac{6.62\times {{10}^{-34}}}{\sqrt{2\times 9.1\times {{10}^{-31}}\times 9.6\times {{10}^{-18}}}} \\\ & \Rightarrow \lambda =1.58\times {{10}^{-10}}m \\\ & \therefore \lambda =1.58\text{ Angstrom} \\\ \end{aligned}
So, the de-Broglie wavelength of an electron beam when accelerated through a potential difference of 60 V will be 1.58 Angstrom.

Hence, Option(A) is correct.

Additional information:
De Broglie equation states that a matter can act as waves just like light which behaves as waves and particles. The equation also explains that a beam of electrons can be diffracted similar to a beam of light. Thus, the de Broglie equation explains the idea of matter having a wavelength. Therefore, according to de Broglie every moving particle whether it is microscopic or macroscopic it will have a wavelength.

Note:
One must remember the value of Planck's constant ‘h’, mass of electron ‘m’ and charge on electron ‘q’, to solve such kinds of questions.
h=6.62×10−34 Joule secondh=6.62\times {{10}^{-34}}\text{ Joule second}
m=9.1×10−31kgm=9.1\times {{10}^{-31}}kg
q=1.6×10−19Cq=1.6\times {{10}^{-19}}C