Find the Width of the Slit Given the Energy of the Electron Beam

[Potatochip911]

Homework Statement

(image is clickable)The figure below shows the distribution of impacts on a screen (see image above) when a beam of electrons of energy $2.5 \cdot 10^{-3}$eV passes through a single slit. The width of the slit is?

Homework Equations

$a\sin\theta=m\lambda \\ I(\theta)=I_{m}(\frac{\sin\alpha}{\alpha})^2 \\ \alpha=\frac{\pi a}{\lambda} \sin\theta \\ I_{0}=1\cdot 10^{-12}W/m^2$

The Attempt at a Solution

[/B]
Since it's a graph of $I/I_{0}$ and at $\theta=0$ the ratio is 1 the maximum intensity is therefore $I_{0}$, It can also be seen that there is a minimum at 1 degree. So in the equation $a\sin\theta=m\lambda$ I am just missing the value for $\lambda$. The intensity equation looks like it won't be solvable for $\alpha$ so I was wondering how to use the fact that they've given the energy passing through the slit. I'm confused as to how I'm supposed to use this information since it doesn't give the number of electrons going through the slit just "a beam of electrons".

 

[ecastro]

Your slit width may be a ratio of physical width length and wavelength.

 

[Potatochip911]

Your slit width may be a ratio of physical width length and wavelength.

I don't understand what you mean by physical width length.

 

[ecastro]

I don't understand what you mean by physical width length.

It is the length of your slit.

 

[ehild]

You need the wavelength of the electrons, and you get it from their momentum; and you get the momentum from the energy.

 

[Potatochip911]

So I have $a=\frac{m\lambda}{\sin\theta}$. Also $\lambda=\frac{h}{m_{e}v}$ but speed is unknown so I tried solving it but I ended up getting the wrong answer. $E=\frac{1}{2}m_{e}v^2$ to solve for speed which results in $a=\frac{mh}{\sin \theta \sqrt{2 E m_{e}}}$ with m=1; sinθ=1°, h=4.14*10^{-15}eV; $m_{e}=9.109*10^{-31}kg$ and E=2.5*10^{-3}eV does not produce the correct answer.

 

[ehild]

Your formula for the wavelength is not correct. The unit of h is not eV.
I suggest to use joules for energy and joule-seconds for h.

 

[Potatochip911]

Your formula for the wavelength is not correct. The unit of h is not eV.
I suggest to use joules for energy and joule-seconds for h.

Okay so I converted to joules but it appears something is wrong with my equation for slit width, if I just plug in the units I get $$a=\frac{(1)(J*s)(m/s)}{(kg)(J)(1)}$$ after cancelling I'm left with $a=\frac{m}{kg}$ which is clearly wrong.

 

[ehild]

The formula $a=\frac{mh}{\sin \theta \sqrt{2 E m_{e}}}$ is wrong. You wrote too many m-s.
What is mv?

 

[Potatochip911]

The formula $a=\frac{mh}{\sin \theta \sqrt{2 E m_{e}}}$ is wrong. You wrote too many m-s.
What is mv?

mv would be the momentum of the electron.
Well this is the oddest question I have ever done, I thought my equation was wrong because I forgot there was a square root when I was checking the units and I could not get the correct answer out of it earlier, but after reviewing the units again at each stage and plugging in the same numbers at the end I got the correct answer. So it turns out that using $a=\frac{mh}{\sin\theta \sqrt{2E\cdot m_{e}}}$ works for using joules and if I want to do the calculation with electron volts I have to use $a=\frac{mh}{\sin\theta \sqrt{\frac{2E\cdot m_{e}}{1.6\cdot10^{-19}}}}$

 

[ehild]

I see; m in the numerator is the order of the minimum, not the mass of the electron.
I usually like to solve a problem symbolically, but that was a case when calculating the numerical value of the speed of the electron and the wavelength of the electron would have been more suggestive and easiest to check. First you need to check if the speed of the electron is not close to the speed of light. The wavelength of the electron should be in the nanometer range. And then, you can get the slit width.Using eV-s unit for h is all right if E and h are in a linear relation as in the case E=hf, for the energy and frequency of a photon. In your formula, h was divided by the square root of energy so you had to work with square root of eV unit, which makes the calculation complicated.