Question

In a double slit experiment, interference is obtained from electron waves produced in an electron gun supplied with voltage $V$. If $\lambda$ is wavelength of the beam, $D$ is the distance of screen, $d$ is the spacing between coherent source, $h$ is Planck’s constant, $e$ is charge on electron and $m$ is mass of electron, then fringe width is given as:
A) $\dfrac{{hD}}{{d\sqrt {2meV} }}$
B) $\dfrac{{2hD}}{{\sqrt {meVd} }}$
C) $\dfrac{{hD}}{{\sqrt {2meVD} }}$
D) $\dfrac{{2hD}}{{\sqrt {meVd} }}$

Answer 1

In this question, we can find the momentum ($p$) of the electron beam by comparing the two different formulas of kinetic energy i.e. $K = eV$and $K = \dfrac{1}{2}\dfrac{{{p^2}}}{m}$. Now, using the formula of wavelength $\lambda = \dfrac{h}{p}$, we can calculate the wavelength ($\lambda$). After finding the wavelength ($\lambda$), we can find the fringe width by using the formula $\beta = \dfrac{{\lambda D}}{d}$.

Complete step by step solution:
According to the question, in a double slit experiment, When the electron has charge $e$ and the voltage of electron waves is $V$ then the kinetic energy $K$ of the electron is given by-
$K = eV$ (i)
But we know that when the mass of the electron is $m$ and velocity is $v$ then the kinetic energy $K$is given as-
$K = \dfrac{1}{2}m{v^2}$ (ii)
Now, comparing the above equation (i) and (ii), we get-
$eV = \dfrac{1}{2}m{v^2}$ (iii)
But the momentum $p$ of the electron having mass $m$ and velocity $v$is given as-
$p = mv$
Or $v = \dfrac{p}{m}$ (iv)
Now, putting the value of $v$ in the equation (iii), we get-
$eV = \dfrac{1}{2}\dfrac{{{p^2}}}{m}$
Or $2meV = {p^2}$
$\Rightarrow p = \sqrt {2meV}$ (v)
If the wavelength of the beam is $\lambda$ and the momentum of the beam is $p$, then
$\lambda = \dfrac{h}{p}$
Or $p = \dfrac{h}{\lambda }$ (vi)
Where $h$ is known as Planck’s constant.
Comparing equation (v) and equation (vi), we get-
$\dfrac{h}{\lambda } = \sqrt {2meV}$
Or $\lambda = \dfrac{h}{{\sqrt {2meV} }}$ (vii)
Now, according to the question, in a double slit experiment, the distance of the screen is $D$ and the spacing between coherent sources is$d$. Then the fringe width $\beta$ is given as-
$\beta = \dfrac{{\lambda D}}{d}$ (viii)
Putting the value of $\lambda$ from equation (vii) in the equation (viii), we get the fringe width-
$\beta = \dfrac{h}{{\sqrt {2meV} }}\dfrac{D}{d}$
Or $\beta = \dfrac{{hD}}{{d\sqrt {2meV} }}$ (ix)
Equation (ix) gives the fringe width of the electron beam.

Hence option (A) is correct.

Note: We can calculate momentum$p$ by comparing $K = eV$and $K = \dfrac{1}{2}\dfrac{{{p^2}}}{m}$ which can be used in the formula of $\lambda = \dfrac{h}{p}$. For finding the fringe width, we need to find the wavelength of the electron beam.