A crack between two walls is $10\text{\hspace{0.17em}cm}$wide. What is the angular width of the central diffraction maximum when

(a) an electron moving at $50\text{\hspace{0.17em}m/s}$passes through?

(b) A baseball of mass$0.145\text{\hspace{0.17em}kg}$ and speed $50\text{\hspace{0.17em}m/s}$passes through?

(c) In each case, an uncertainty in momentum is introduced by the “experiment” (i.e., passing through the slit). Specifically, what aspect of the momentum becomes uncertain, and how does this uncertainty compare with the initial momentum of each?

The width of the crack between two walls is $10\text{\hspace{0.17em}cm}$.The speed of the electron and the baseball is $50\text{\hspace{0.17em}m/s}$, and the weight of the baseball is$0.145\text{\hspace{0.17em}kg}$ .

It is known that the diffraction minima occur at $m\lambda =Dsin\theta$.

Here m is integer and D is the slit width, $\mathbf{\theta }$is the angle made by the incident wave with the normal and$\mathbf{\lambda }$ is wave length of the incident wave.

(a)

The diffraction minima for a single slit have the relation,

$m\lambda =D\sin \theta$.

The value of is $1$,so, the full width can be obtained using the formula:

$\Delta \theta =2{\sin }^{-1}\left(\frac{\lambda }{D}\right)$

Now, substitute $h=6.63\times {10}^{-34}{\text{\hspace{0.17em}m}}^{\text{2}}\cdot \text{kg/s}$and $m=9.11\times {10}^{-31}\text{\hspace{0.17em}kg}$for electron and $v=50\text{\hspace{0.17em}m/s}$to find the value of $\lambda$by using the de-Broglie relation as:

$\begin{array}{c}\lambda =\frac{h}{mv}\\ =\frac{6.63\times {10}^{-34}{\text{\hspace{0.17em}m}}^{\text{2}}\cdot \text{kg/s}}{9.11\times {10}^{-31}\text{\hspace{0.17em}kg}\times 50\text{\hspace{0.17em}m/s}}\\ =1.46\times {10}^{-5}\text{\hspace{0.17em}m}\end{array}$

Substitute $\lambda =1.46\times {10}^{-5}\text{\hspace{0.17em}m}$and$D=10\text{\hspace{0.17em}cm}=0.1\text{\hspace{0.17em}m}$ to find the spreading as follows:

$\begin{array}{c}\Delta \theta =2{\sin }^{-1}\left(\frac{\lambda }{D}\right)\\ =2{\sin }^{-1}\left(\frac{1.46\times {10}^{-5}\text{\hspace{0.17em}m}}{0.1\text{\hspace{0.17em}m}}\right)\\ =2\times 0.00835°\\ =0.0167°\end{array}$

In this case, the electron is able to produce a diffraction pattern that is at a very small angle but measurable.

Thus, The angular width of the central diffraction maximum when an electron passes through is $0.0167°$.

(b)

Substitute $h=6.63\times {10}^{-34}{\text{\hspace{0.17em}m}}^{\text{2}}\cdot \text{kg/s}$and $m=0.145\text{\hspace{0.17em}kg}$for electron and $v=50\text{\hspace{0.17em}m/s}$to find the value of $\lambda$by using the de-Broglie relation:

$\begin{array}{c}\lambda =\frac{h}{mv}\\ =\frac{6.63\times {10}^{-34}{\text{\hspace{0.17em}m}}^{\text{2}}\cdot \text{kg/s}}{0.145\text{\hspace{0.17em}kg}\times 50\text{\hspace{0.17em}m/s}}\\ =9.14\times {10}^{-35}\text{\hspace{0.17em}m}\end{array}$

Substitute $\lambda =9.14\times {10}^{-35}\text{\hspace{0.17em}m}$and$D=10\text{\hspace{0.17em}cm}=0.1\text{\hspace{0.17em}m}$ to find the spreading as follows:

$\begin{array}{c}\Delta \theta =2{\sin }^{-1}\left(\frac{\lambda }{D}\right)\\ =2\times {\sin }^{-1}\left(\frac{9.14\times {10}^{-35}\text{\hspace{0.17em}m}}{0.1\text{\hspace{0.17em}m}}\right)\\ =2\times 5.24\times {10}^{-32}°\\ =1.05\times {10}^{-31}°\end{array}$

This is a very small angle. Even Labs won’t be able to detect this interference pattern.

Thus, the angular width of the central diffraction maximum when baseball passes through is$1.05\times {10}^{-31}°$ .

(c)

Le the particle directed along the $x$-direction with the slit opening along the$y$-direction. This wave confinement along $y$-direction will produce uncertainty in the momentum $\Delta p_y$. We have to find the uncertainty in the $y$-component of momentum using the Heisenberg relation.

The first case is of the electron:

$\begin{array}{c}\Delta y\Delta p_y\ge \frac{\hslash }{2}\\ \Delta p_y\ge \frac{\hslash }{2\Delta y}\\ \Delta p_y\ge \frac{1.054\times {10}^{-34}{\text{\hspace{0.17em}m}}^{\text{2}}\cdot \text{kg/s}}{2\times 0.1\text{\hspace{0.17em}m}}\\ =5.27\times {10}^{-34}\text{\hspace{0.17em}kg}\cdot \text{m/s}\end{array}$

The initial momentum can be caculated as,

$\begin{array}{c}{p}_i=p_x\\ =mv\\ =9.11\times {10}^{-31}\text{\hspace{0.17em}kg}\times 50\text{\hspace{0.17em}m/s}\\ =4.55\times {10}^{-29}\text{\hspace{0.17em}kg}\cdot \text{m/s}\end{array}$

So, the value of $\frac{\Delta p_y}{p_i}$can be caculated as,

$\begin{array}{c}\frac{\Delta p_y}{p_i}=\frac{5.27\times {10}^{-34}\text{\hspace{0.17em}kg}\cdot \text{m/s}}{4.55\times {10}^{-29}\text{\hspace{0.17em}kg}\cdot \text{m/s}}\\ =1.16\times {10}^{-5}\end{array}$

The second case for baseball:

$\begin{array}{c}\Delta y\Delta p_y\ge \frac{\hslash }{2}\\ \Delta p_y\ge \frac{\hslash }{2\Delta y}\\ \Delta p_y\ge \frac{1.054\times {10}^{-34}{\text{\hspace{0.17em}m}}^{\text{2}}\cdot \text{kg/s}}{2\times 0.1\text{\hspace{0.17em}m}}\\ =5.27\times {10}^{-34}\text{\hspace{0.17em}kg}\cdot \text{m/s}\end{array}$

The initial momentum can be caculated as,

$\begin{array}{c}{p}_i=p_x\\ =mv\\ =0.145\text{\hspace{0.17em}kg}\times 50\text{\hspace{0.17em}m/s}\\ =7.25\text{\hspace{0.17em}kg}\cdot \text{m/s}\end{array}$

So, the value of $\frac{\Delta p_y}{p_i}$can be caculated as,

$\begin{array}{c}\frac{\Delta p_y}{p_i}=\frac{5.27\times {10}^{-34}\text{\hspace{0.17em}kg}\cdot \text{m/s}}{7.25\text{\hspace{0.17em}kg}\cdot \text{m/s}}\\ =7.27\times {10}^{-35}\end{array}$

From the above calculation, the ratio of uncertainty and the initial momentum is small but measurable in the case of the electron. Still, the ratio is very small and negligible in calculations.

Thus, the longer the wavelength of the object, the higher the disturbance associated with it (whether angular dispersion or momentum uncertainty).