91Ó°ÊÓ

It is known that the diffraction minimum can be obtained at $\mathbf{sin}{\mathit{θ}}_{\mathbf{1}}\mathbf{=}\frac{\mathbf{λ}}{\mathbf{w}}$.

From the graph, it is clear that the wavelength is equal to 1 nm. So, the momentum can be obtained as follows:

$\begin{array}{l}p=\frac{h}{\mathrm{λ}}\\ =\frac{h}{1.0×{10}^{-9}m}\\ =\left({10}^9{m}^{-1}\right)h\end{array}$

Thus, the momentum is $\left({10}^9{m}^{-1}\right)h$.

From the graph, it is clear that the peak lies at $k=2\mathrm{Ï€}{\mathrm{nm}}^{-1}$. So, the momentum can be obtained as follows:

$$\begin{gathered} p=\mathrm{Ä\S }k \\ =\frac{h}{2\mathrm{Ï€}}×2\mathrm{Ï€}×{10}^9{\mathrm{m}}^{-1} \\ =\left({10}^9{\mathrm{m}}^{-1}\right)\mathrm{h} \end{gathered}$$

Since, the momentum is same as the previous part. So, it demonstrates the same thing and related by the uncertainty principle.

Here, the peaked value of the momentum is equal to$({10}^9{\mathrm{m}}^{-1})h$. So, the momentum will be most likely found closest to peaked value.

So, accordingly$(0.9×{10}^9{m}^{-1})h$and $\left(1.1×{10}^9{m}^{-1}\right)h$will be most likely values.

By the same reason,$\left(0.5×{10}^9{m}^{-1}\right)h$will be less likely values because it away from the peak value (half the peak value).

Thus,$(0.9×{10}^9{m}^{-1})h$and $(1.1×{10}^9{m}^{-1})h$would be likely values and $(0.5×{10}^9{\mathrm{m}}^1)h$would be unlikely value.

The two main rules are considered while drawing transformed graph:

An inverse relationship between the confinement in the position, momentum, or wave-number is used to express the uncertainty relation. As a result, the wave's momentum is more tightly contained the more spreading it has in its location, and vice versa.

Since the relationship between the wavenumber and wavelength$\left(k=\frac{2\mathrm{π}}{\mathrm{λ}}\right)$is inverse, Consequently, the value of the wave numberr in the modified graph is less the longer the wave's wavelength (where its centred around in our case here).

So, the graph of${\mathrm{ψ}}_1\left(x\right)$will have less spreading. The spreading in k-space will be more and the peaked value be same. The graph of${\mathrm{ψ}}_2\left(x\right)$has wavelength two times to$\mathrm{ψ}$graph. So, the peaked value occurs at half the value of$\mathrm{ψ}$graph that is$\mathrm{π}{\mathrm{nm}}^{-1}$.

The Fourier transformation sketch is given below: