The neutrons in a parallel beam, each having kinetic energy 1/40 eV, are directed through two slits 0.36 mm?

Question

apart. How far apart will the interference peaks be on a screen 1.1 m away? [Hint: first find the wavelength of the neutron.]

Answer 1

E = hc / lamda for massless particles only like photons.

For massive particles
E = p^2 / 2m = (k hbar)^2 / 2m = (h/lambda)^2 / 2m
lambda ^2 = h^2 / (2mE)

lamda = h sqrt( 1/(2mE))
h = 4.136 e-15 eV s
m (neutron) = 0.940e9 eV/c^2
c = 3e8 m/s

Finally:
lamda = 4.136 e-15 eV s x 3e8 m/s x 4.61 e-3 / eV =
0.572e-8 m

Answer:
distance = (lambda / h) d = (0.572e-8 m/ 0.36e-3 m) * 1.1 m = 1.75 e -5 m = 17.5 micron

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Edit:
Uhmm, I screwed up arithemetic as usually.
lambda = 0.572e-8 m = 57.2A is too much for neutron
at such high energies (about room temperature).

Correction:

lamda = 4.136 e-15 eV s x 3e8 m/s x 0.1458 e-3 / eV =
= 0.181e-9 m
= 1.81 Angstroem

Answer:
distance = (lambda / h) d = (0.181e-9 m/ 0.36e-3 m) * 1.1 m = 0.553 e-6 m
= 0.553 micron

Answer 2

energy of neutrons E= 1/40 * 1.6*10^-19 joules
E= 4*10^-21 joules

E = hc / lamda
lamda = hc / E = 6.6*10^-34*3*10^8/4*10^-21 meter
lamda = 4.95*10^-5 meter


x-bar (width) = D* lamda / d

where D = 1.1 m and d = 0.36/1000 m

x-bar (width) = 1.1*4.95*10^-5 /0.36*10^-3
=15.125 *10^-2 meter
= 0.15125 m