De Broglie's Wavelength (matter wave)

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wavingerwin
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Homework Statement


Parallel beam of "cold" neutrons in the x-axis, with energy 0.1eV strikes a slit system in y-z plane.

Slits are parallel to the y-axis 1 micron apart. What will be the distance between the detector at x=20m which records the highest flux of neutrons, and the closest one to it that records nearly zero flux?

Homework Equations


lambda = h/p
f = E/h

The Attempt at a Solution


I tried to find lambda
going with

lambda = h/p
p = gamma*m*v

v = c*(1-(K/mc2+1)-2)0.5

with K = 0.1eV = 1.602e-20 Joules
m = 1.675e-27
I found v=0... since K2/mc2 << 1

What am I doing wrong?
Please help...

Thank you in advance
 
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I see :smile:.

So now that I have lambda = 9.04e-11 m

Can I use the equation (n-0.5)lambda = d sin(theta)

with n = 1

and x = 20 tan(theta)

so that x is the closest distance between the central maxima and the area of nearly zero flux?

I am concerned since the problem states that the neutrons go through a "slit system" and my tutor confirmed that this means that it has multiple slits, can I still use (n-0.5)lambda = d sin(theta) which I understand is mainly used for dual-slit?

if not, what equation should I use?

Since it says that the neutrons are in a "beam", can I say that even though there are multiple slits (1 micron apart), only 2 slits are playing a part in the problem?

thank you
 
:shy:...It does not do anything...

I get it now that the 'slit system' is juts a diffraction grating and reading back my old notes about diffraction grating, the only difference is that now I cannot approximate sin(theta)=tan(theta)...

sorry about my silliness..
 
Thanks vela..

the last part of the question says:
If the neutron come originally from a U-235 nuclear fission reactor, in what sense are they now "cold"?

I looked at wikipedia, and found "The neutron temperature, also called the neutron energy, indicates a free neutron's kinetic energy"

My guess is the neutrons are cold in the sense that they are traveling 'slow'.

Would that be right?

Again, thanks you for your time & patience in advance. :smile: