Question

Integrated Concepts

(a) What is the separation between double slits that produces a second order minimum at 45.0^o for 650 - nm light?

(b) What slit separation is needed to produce the same pattern for 1.00 keV protons.

Short answer

(a)The separation between double slits that produces a second order minimum at 45.0^o for 650 - nm light is 2.30 x 10^-6 m.

(b)The separation needed to produce the same pattern for 1.00 keV protons is 3.2 x 10^-12 m.

Step-by-step solution

Determine the formulas:

Consider the equation for the separation.

$$\mathbf{d}\mathbf{s}\mathbf{i}\mathbf{n}\mathbf{(}\theta \mathbf{)}\mathbf{=}\left(\mathbf{m}\mathbf{+}\frac{\mathbf{1}}{\mathbf{2}}\right)\lambda$$

In the equation 29.23, the momentum is given by the equation:

$$\rho \mathbf{=}\frac{\mathbf{h}}{\lambda }$$

Here the value of $$\mathrm{h}=6.626\times 10^{-34}\mathrm{J}\mathrm{s}$$is said to be the Planck's constant.

The value of λ is said to be the photon wavelength.

By solving the first equation, the wavelength of the value λ is:

$$\lambda =\frac{h}{p}$$

In the equation 29.26, electron has kinetic energy, which is classically obtained using:

$$\mathbf{K}{\mathbf{E}}_{\mathbf{p}}\mathbf{=}\frac{\mathbf{1}}{\mathbf{2}}\mathbf{m}{\mathbf{v}}^{\mathbf{2}}$$

Here, the value of mis said to be the mass of the electron.

The value of vis said to be the velocity of the electron.

Multiply and divide the fifth equation with mass of the value m as follows:

$$\begin{array}{c}K{E}_p=\frac{1}{2}m{v}^2\frac{m}{m}\\ =\frac{1}{2m}{\left(mv\right)}^2\\ =\frac{{\left(mv\right)}^2}{2m}\end{array}$$

Rewrite the equation as:

$$K{E}_p=\frac{p^2}{2m}$$

Rewrite the expression in terms of the momentum as:

$$\mathbf{p}\mathbf{=}\sqrt{\mathbf{2}\mathbf{m}\mathbf{K}{\mathbf{E}}_{\mathbf{e}}}$$

Find separation between Double slits 

(a)

Use m = 2 in the equation for separation and solve as:

$$d=\frac{\left(m+\frac{1}{2}\right)\lambda }{\sin (\theta )}$$

$$=2.30\times {10}^{-6}\mathrm{m}$$

Hence, the separation between double slits that produces a second order minimum at 45.0^o for 650 - nm light is 2.30 x 10^-6 m.

Find separation for 1-kev

(b)

Derive the expression for the momentum as follows:

$$\begin{array}{c}p=\sqrt{2mK.E}\lambda \\ =\frac{h}{\sqrt{2mK.E}}d\\ =\frac{\left(m+\frac{1}{2}\right)\lambda }{\sin (\theta )}\\ =\frac{\left(m+\frac{1}{2}\right)h}{\sin (\theta )\sqrt{2mK.E}}\end{array}$$

Substitute the values and solve as:

$$=3.2\times {10}^{-12}\mathrm{m}$$

Therefore, the separation needed to produce the same pattern for 1.00 keV protons is 3.2 x 10^-12 m.