Question

The width of the central maximum is defined as the distance between the two minima closest to the center of the diffraction pattern. Since these are symmetric about the center of the pattern, you need to find only the distance to one of the minima, and then the width of the central maximum will be twice that distance. Find the angle θ between the center of the diffraction pattern and the first minimum. The equations for diffraction, which you have seen applied to light, are valid for any wave, including electron waves. Recall that the angle to a diffraction minimum for single-slit diffraction is given by the equation sin(θ)=mλ/a, where a is the width of the slit and m is an integer. Recall that m=±1 for the first minima on either side of the central maximum. Do not make any approximations at this stage. Express your answer in terms of h, a, me, e, and V.

Answer 1

Answer:

The value of the angle is $\bf{ \sin^{-1}[h/am_{e}v]}$.

Explanation:

Given:

The condition for diffraction minima is

$a \sin \theta = m \lambda~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~(1)$

where, $a$ is the slit-width, $\theta$ is the angle of incidence, $m$ is the order number and $\lambda$ is the wavelength of the light.

The wavelength of an electron traveling through a medium is governed by de Broglie's hypothesis.

According to de Broglie's hypothesis

$\lambda &=& \dfrac{h}{p}\\ &=& \dfrac{h}{m_{e}v}$

Here, $h$ is Planck's constant, $m_{e}$ is the mass of the electron and $v$ is the velocity of the electron.

For first minimum $m = 1$.

From equation (1), we have

$&& a \sin \theta = \dfrac{h}{m_{e}v}\\&or,& \theta = \sin^{-1}[\dfrac{h}{am_{e}v}]$