Question

Electrons are accelerated through a 20 V potential difference producing a monoenergetic beam. This is directed at a double-slit apparatus of 0.010 mm slit separation. A bank of electron detectors is 10 m beyond the double slit. With slit 1 alone open, 100 electrons per second are detected at all detectors. With slit 2 alone open, 900 electrons per second are detected at all detectors. Now both slits are open.

(a) The first minimum in the electron count occurs at detector X. How far is it from the center of the interference pattern?

(b) How many electrons per second will be detected at the center detector?

(c) How many electrons per second will be detected at detector X?

Short answer

(a)

Center of the interference pattern$\mathbf{\text{0.14mm}}$

(b)

For the constructive interference square amplitude at the center detector, X is${\left|{\text{ψ}}_T\right|}^{\text{2}}\propto 16\text{00}$, which means 1600 electrons per second.

(c)

For the destructive interference square amplitude at the center detector is X${\left|{\text{ψ}}_D\right|}^{\text{2}}\propto 4\text{00}$, which means 900 electrons per second.

Step-by-step solution

Given.

$\text{V=20V}$

$D=10\text{m}$

${\text{d=10}}^{\text{-5}}\text{m}$

${\left|{\text{ψ}}_{\text{1}}\right|}^{\text{2}}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}α\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}100}$

${\left|{\psi }_2\right|}^2\alpha 900$

Step 2:Concept Introduction

We know that when a charged particle of charge q and mass m is accelerated with a potential, $\mathbf{\text{V}}$, the kinetic energy$\mathbf{\left(}\frac{\mathbf{\text{1}}}{\mathbf{\text{2}}}{\mathbf{\text{mv}}}^{\mathbf{\text{2}}}\mathbf{\right)}$can be expressed as,

$\mathbf{\text{qV=}}\frac{\mathbf{\text{1}}}{\mathbf{\text{2}}}{\mathbf{\text{mv}}}^{\mathbf{\text{2}}}$……………………(1)

and that the velocity, v, is proportional to the wavelength using the de Broglie formula.

$\mathbf{\text{v=}}\frac{\mathbf{\text{p}}}{\mathbf{\text{m}}}\mathbf{\text{=}}\frac{\mathbf{\text{h}}}{\mathbf{\text{mλ}}}$………..…………(2)

Calculations for λ

We need to eliminate v by using the equations (1) and (2) to find the wavelength$\text{λ}$of the electron.

First, we must describe both equations in terms of ${\text{v}}^{\text{2}}$, such that

$\frac{\text{2qV}}{\text{m}}{\text{=v}}^{\text{2}}$

${\text{v}}^{\text{2}}\text{=}\frac{{\text{h}}^{\text{2}}}{{\text{m}}^{\text{2}}{\text{λ}}^{\text{2}}}$

Then, we combine both equations.

$\frac{\text{2qV}}{\text{m}}\text{=}\frac{{\text{h}}^{\text{2}}}{{\text{m}}^{\text{2}}{\text{λ}}^{\text{2}}}$

$\text{2qV=}\frac{{\text{h}}^{\text{2}}}{{\text{mλ}}^{\text{2}}}$

Lastly, we isolate$\text{λ}$

$\text{λ=}\frac{\text{h}}{\sqrt{\text{2qVm}}}$……………………..(3)

Now, substitute the given parameters in equation (3) to obtain the wavelength of the electron,

$\begin{array}{c}\text{λ}=\frac{\text{6.63}\times {\text{10}}^{\text{-34}}J\cdot s}{\sqrt{\text{2}\times (\text{1.6}\times {\text{10}}^{\text{-19}}C)\times \text{(20\hspace{0.17em}V)}\times (\text{9.1}\times {\text{10}}^{\text{-31}}kg)}}\\ =\text{2.75}\times {\text{10}}^{\text{-10}}\text{m}\end{array}$

Thus the wavelength of the electron is${\text{2.75×10}}^{\text{-10}}\text{\hspace{0.17em}m}$

Interference equation.

(a)

Using the constructive interference equation, we get,

$\text{mλ=dsin(θ)}$

Using sine law, we get,

$\frac{\text{1}}{\text{2}}\text{λ=d}\frac{\text{y}}{\text{D}}$

Hence, substituting the given information in the above equation, we get,

$\begin{array}{c}\text{y}=\frac{\text{D}}{\text{2d}}\text{λ}\\ =\frac{\text{10}}{\text{2}\times \left({\text{10}}^{\text{-5}}m\right)}\times (\text{2.75}\times {\text{10}}^{\text{-10}}m)\\ =0.14mm\end{array}$

Hence, the distance of the first minimum from the center is$\text{0.14mm}$

Constructive interference.

(b)

To determine how many electrons per second are at the center of the detector, we solve using constructive interference.

For slit 1,

$\begin{array}{l}{\left|{\text{ψ}}_{\text{1}}\right|}^{\text{2}}\propto \text{100}\\ \left|{\text{ψ}}_{\text{1}}\right|\propto \text{10}\end{array}$……………..(4)

For slit 2,

$\begin{array}{l}{\left|{\text{ψ}}_2\right|}^{\text{2}}\propto 9\text{00}\\ \left|{\text{ψ}}_2\right|\propto 3\text{0}\end{array}$…………………….(5)

Therefore, combining equations (4) and (5) for the constructive interference, we get the output of

$\begin{array}{c}\left|{\text{ψ}}_{\text{T}}\right|\propto \left|{\text{ψ}}_1\right|+\left|{\text{ψ}}_2\right|\\ \propto \text{10}+\text{30}\\ \propto \text{40}\end{array}$

Therefore,

${\left|{\text{ψ}}_T\right|}^{\text{2}}\propto 16\text{00}$

Destructive Interference.

(c)

To determine how many electrons per second are at detector X, we use equations (4) and (5) for the destructive interference, such that,

$\begin{array}{c}\left|{\text{ψ}}_{\text{D}}\right|\propto \left|{\text{ψ}}_2\right|-\left|{\text{ψ}}_1\right|\\ \propto \text{30}-1\text{0}\\ \propto \text{20}\end{array}$

Therefore,

${\left|{\text{ψ}}_{\text{D}}\right|}^2\propto \text{400}$