Question

Calculate the accelerating potential that must be imparted to a proton beam to wavelength of$\text{0}\text{.005 nm}$.

Answer 1

Hint: The French physicist de Broglie suggested that light has dual properties of both particle and wave, the particle nature of electron such electron, proton also have the property of waves. He derive the wavelength of such particle -$\text{ }\!\!\lambda\!\!\text{ = }\dfrac{\text{h}}{\text{mv}}.......\text{(i)}$
Where these letters have their usual meanings. Where $\text{h}$ Represents the Planck constant and its value are$\text{6}\text{.63 }\!\!\times\!\!\text{ 1}{{\text{0}}^{\text{-34}}}\text{Js}$, $\text{m}$and $\text{v}$represent the mass and velocity respectively.
Consider a proton beam of mass $\text{m}$and charge$\text{+e}$. Let$\text{v}$is the final velocity attained by the proton beam when it is accelerated from rest through a potential difference$\text{V}$. Then kinetic energy gain by the proton will be equal to the work done on the proton by the electric field. So kinetic energy gained by proton will be $\text{K=}\,\dfrac{\text{1}}{\text{2}}\text{m}{{\text{v}}^{\text{2}}}\,\text{=}\,\dfrac{{{\text{P}}^{\text{2}}}}{\text{2m}}$
And work done by the electric field on the proton = $\text{eV}$
Since work done = kinetic energy so, $\text{K = }\,\text{eV}......\text{(ii)}$

Complete step by step solution:
We will calculate the accelerating potential in two steps:
In the first step we will calculate the velocity of the proton beam by applying the equation (i).
$\begin{align}
& \text{ }\!\!\lambda\!\!\text{ = }\dfrac{\text{h}}{\text{mv}}.......\text{(i)} \\
& \text{v =}\,\,\dfrac{\text{h}}{\text{m }\!\!\lambda\!\!\text{ }} \\
& \,\,\,=\,\,\dfrac{6.63\times \,{{10}^{-34}}\text{Js}}{(1.67\times {{10}^{-27}}\text{Kg)(0}\text{.005}\times \text{1}{{\text{0}}^{-9}}\text{m})} \\
& \text{v}=\,7.94\times {{10}^{4}}{\scriptstyle{}^{\text{m}}/{}_{\text{s}}} \\
\end{align}$
In second step we will calculate kinetic energy of proton
\[\begin{align}
& \text{K}=\dfrac{1}{2}m{{v}^{2}}\, \\
& \,\,\,\,\,=\,\dfrac{1}{2}(1.67\times {{10}^{-27}}\text{Kg}){{(7.94\times {{10}^{4}}{\scriptstyle{}^{\text{m}}/{}_{\text{s}}})}^{2}} \\
& \text{K}=\,5.26\times {{10}^{-18}}\text{J} \\
& \text{K =}\,\dfrac{5.26\times {{10}^{-18}}}{1.602\times {{10}^{-19}}}\text{eV} \\
& \text{K}\,\text{=}\,\,32.8\,\text{eV} \\
\end{align}\]
By the application of equation (ii) we get.
As the magnitude of charge of a proton is the same as that of an electron, the potential required is equal in magnitude to the number of $\text{eV}$. So the required potential will be\[32.8\].

Note: Magnitude of charge of electron is equal to the magnitude of charge of proton.
Mass of the proton must be kg because the plank constant is present in the S.I unit.
When a charged particle carrying Q coulomb is accelerated by applying potential difference V then kinetic energy will be the product of charge and applied potential.