Chapter 28, Problem 81P

To determine

The slit width.

Answer to Problem 81P

The slit width is $352\times {10}^{-9}\text{m}$.

Explanation of Solution

The width of the central maximum is $\text{1}\text{.13}\text{mm}$, slit-screen distance is $1.0\text{m}$ and the accelerating potential difference is $38.0\text{ V}$.

Write the expression for de Broglie wavelength

$\lambda =\frac{h}{mv}$                                           (I)

Here, $\lambda$ is the de Broglie wavelength, $h$ is the Planck’s constant, $m$ is the mass of the electron and $v$ is its velocity.

Write the expression for kinetic energy

$K=\frac{1}{2}m{v}^2$                                           (II)

Here, $K$ is the kinetic energy of the electron.

Rearranging the above equation

$v=\sqrt{\frac{2K}{m}}$                                           (III)

Substituting (III) in (I)

$\lambda =\frac{h}{m}\sqrt{\frac{m}{2K}}$

$\lambda =\frac{h}{\sqrt{2mK}}$                                           (IV)

Write the expression for kinetic energy in terms of accelerating potential

$K=eV$                                            (V)

Here $e$ is the charge of the electron and $V$ is the accelerating potential difference.

Substituting (V) in (VI)

$\lambda =\frac{h}{\sqrt{2meV}}$                                           (VI)

Write the expression for first diffraction minima

$\sin \theta =\frac{\lambda }{a}$                                                                    (VII)

Here, $\theta$ is the angle of first diffraction minima, $\lambda$ is the wavelength and $a$ is the slit width.

Using small angle approximation i.e. $\sin \theta \approx \theta$ in (VII) and rearranging

$\lambda =\theta a$                                                                    (VIII)

Write the angle of diffraction (refer figure 1)

$\tan \theta =\frac{x}{D}$                                                                     (IX)

Here, $x$ is half the width of the central maximum, $D$ is the slit-screen distance

Using small angle approximation i.e. $\tan \theta \approx \theta$ in (IX)

$\theta =\frac{x}{D}$                                                                     (X)

Substituting (X) in (VIII)

$\lambda =\frac{x}{D}a$                                                                    (XI)

Equating (IV) and (IX)

$\frac{h}{\sqrt{2meV}}=\frac{x}{D}a$

Rearranging for $a$

  $a=\frac{hD}{x\sqrt{2meV}}$                                                                 (XII)

Substitute $9.109\times {10}^{-31}\text{ kg}$ for $m$, $1.00\text{ m}$ for $D$, $38\text{ V}$ for $V$, $1.13\text{mm}/2$ for $x$ and $6.626\times {10}^{-34}\text{ Js}$ for $h$ in (XII) to find $a$

$\begin{array}{c}a=\frac{6.626\times {10}^{-34}\text{ Js}\times 1.00\text{ m}}{\left(1.13\text{mm}/2\right)\sqrt{2\times 9.109\times {10}^{-31}\text{ kg}\times 38.0\text{ eV}}}\\ =\frac{6.626\times {10}^{-34}\text{ Js}\times 1.00\text{ m}}{\left(0.565\times {10}^{-3}\text{m}\right)\sqrt{2\times 9.109\times {10}^{-31}\text{ kg}\times 38.0\text{ eV}\times 1.602\times {10}^{-19}\text{ J}}}\\ =352\times {10}^{-9}\text{m}\\ =352\text{nm}\end{array}$

Thus, the slit width is $352\times {10}^{-9}\text{m}$.