# Quantum physics homework

1. Dec 1, 2013

### pixiv

1. The problem statement, all variables and given/known data

Consider a two-slit experiment using electrons. The slits are assumed to be very narrow compared to the wave-length of the electrons. Beyond the slits is a bank of electron detectors. At the center detector, directly in the path the beam would follow if unobstructed, 100 electrons per second are detected. Suppose that as the detector angle varies, the number per unit time of electron’s arriving varies from a maximum of 100/s to a minimum of 0. Suppose the electrons have K = 1.0 eV of kinetic energy and the narrow slits are separated by 0.020 μm.
a/ At what angle is the detector X located where the minimum is reached ?
b/ How many electrons would be detected per second at the center detector if one of the slits were blocked?
c/ How many electrons would be detected per second at the center detector and at detector X if one of the slits were narrowed to 36% of its original width ?

2. Relevant equations

3. The attempt at a solution
a/
Someone told me to use the Brag equation: 2dSinα=mλ
I don't understand why I have to use it in this problem.
But if I use this equation:
I have d=0.020.020 μm ; m=1 (minimum);
K->v->p->λ=$\frac{h}{p}$
So I can figure out the α
b/
2 split:
|ψ|2≈100/s => |ψ|≈10/s
=> 1 split |ψ|≈5/s => |ψ|2≈25/s

c/
* At the center detector:
the narrow split ψ1:
=> |ψ1|2≈25/s*0.36=9/s
=> |ψ1|=3/s
=> the total=|ψ1|+|ψ2|=5+3
=> |ψT|2≈64/s
* At detector X:
1|-|ψ2|=5-3=2/s
=> |ψT|2≈4/s

I'm not sure that my solution is right. Please, help me to check it. Thank you very much.