# Homework Help: Two slit diffraction, neutron, kinetic energy

1. Apr 4, 2009

### Yroyathon

hi again folks. this week's set of problems is getting my goat. here's another I can't seem to figure out.

1. The problem statement, all variables and given/known data
In a neutron two-slit diffraction experiment, the slits are 65 µm apart. If the third diffraction maximum is detected at an angle of 2 multiplied by 10-7 rad, what is the kinetic energy of the neutrons?

2. Relevant equations
E=h*f=(h*c)/lambda
d * sin theta_n = n * lambda

3. The attempt at a solution

I set d = 65 * 10^(-6), and n=3. theta_n = 2*10^(-7). I used this to solve for lambda. Given the energy equation, I computed (h*c)/lambda to get 4.59 * 10^(-14) eV, but this is wrong. I thought I had a clear application of this equation here, but there has to be something I'm not quite getting that's different, that I'm not taking into account here.

can anyone see what I've missed?

I'd appreciate any ideas you might have.

,Yroyathon

2. Apr 4, 2009

### rl.bhat

You are using neutron, not photon.

3. Apr 5, 2009

### Yroyathon

oh. ok. so instead of using E=h/f, should I substitute Broglie wavelength lambda=h/p, planck's constant divided by momentum?

and then use p=sqrt(2*m*E), momentum equals the square root of the product of 2, the mass of the neutron, and its kinetic energy? using this and the original d * Sin[theta_n] = n * lambda, I believe I can find the kinetic energy.

does this sound right?

4. Apr 5, 2009

### Yroyathon

well, that approach didn't work. is the equation p=sqrt(2mE) not valid for this scenario?

5. Apr 4, 2010

### Physics321

Yes, I would really like some additional input on this problem as well, if anyone has anything to offer.

I have tried all of the steps Yroyathon has done and still have not arrived at the correct answer.

Except what I did was do (6.5E-5sin(2^-7))/(3)=lambda.

From this I used K= (h^2)/2m(lambda^2) and plugged everything in like so

K=(6.63E-34)^2/(2*1.6749E-27 kg)(4.33E-12)^2, which equals 6.98E-18, then i used the analogy 1 eV= 1.6E-19 J, so then took (6.98E-18/1.6E-19)= 4.367 eV.

I'm not sure what Yroyathon and I are doing wrong?

6. Apr 4, 2010

### ideasrule

6.98E-18/1.6E-19 is 43.67, not 4.367.

7. Apr 5, 2010

### 1989.1989

despite this miscalculation, the answer is still wrong.
there is something wrong with either the solution or the question.

8. Apr 5, 2010

### rl.bhat

In the diffraction pattern, the third maximum is obtained when the path difference is 5λ/2, assuming that the central maximum is the first maximum.

9. Apr 5, 2010

### Physics321

In a neutron two-slit diffraction experiment, the slits are 100 µm apart. If the third diffraction maximum is located at an angle of 2.0x〖10〗^(-7) rad, what is the kinetic energy of the neutrons?

Relevant equations:
nλ=Dsinθ
K=p^2/2m=h^2/(2mλ^2 )

Thus we have then
λ=((100x〖10〗^(-6) m) sin⁡(2.0x〖10〗^(-7) rad))/3=6.67x〖10〗^(-12) m

The kinetic energy then for a low-energy neutron is as follows:
K=(6.63x〖10〗^(-34) J∙s)^2/[2(1.67x〖10〗^(-27) kg) (6.67x〖10〗^(-12) m)^2 ] = 2.958x〖10〗^(-18) J.

Finally, if you would like to convert it to eV, then:
Recall 1 eV=1.6x〖10〗^(-19) J
(2.958x〖10〗^(-18) J)/(1.6x〖10〗^(-19) J)=18.4 eV ≈19 eV

Please don’t correct my minor mistakes such as the number of decimals or a letter missing. If you would like to explain anything then conceptually explain it.

10. Apr 5, 2010

### Cilabitaon

It's actually 43.625...

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