Electrons emerge from an electron gun with a speed of 2.0 times 10^6 m/s (with t

Question

Electrons emerge from an electron gun with a speed of 2.0 times 10^6 m/s (with this speed, it is OK to treat the electron as a non-relativistic particle) and then pass through a pair of thin parallel slits. Interference fringes (close to the center of the pattern) with a spacing of 2.7 mm are detected on a screen very far away from the double slits. (a) What is the de Broglie wavelength for the electrons? (b) you forgot to measure the distance from the slits to the screen, but for sure you remember it to be more than 1.0 m You are asked to predict what would the fringe spacing be if the electrons were replaced by protons with the same kinetic energy (without knowing the slits' separation and the distance from the slits to the screen)? (c) you now went back and found that the distance from the slits and the screen is 2.0 m. What is the slits' separation?

Explanation / Answer

mass of electron = Me = 9.1*10^-31 kg
speed of electron = v = 2*10^6 m/s
intererence spacing, d = 2.7 mm = 2.7*10^-3 m

a) Debroglie wavelength, lambda = h/mv = 6.626*10^-34 /9.1*10^-31* 2*10^6 = 3.640*10^-10 m [ h = planks constant, m is mas of electron and v is speed of electron]
b) fringe spacing with small angle sapproximation, d = lambda*D/a [ where D is distance from the screen, a is the slit width]
so, for protoins, fringe spaceing, d' = lambda'*D/a
d' = d*(lambda'/lambda) = d*(h/Mp*u) / (h / Me*v) (Mp is mass of proton)
u is velocity of proton
now for same KE
Mp*u^2 = Me*v^2
d' = d*(lambda'/lambda) = d*(h/Mp*v*sqroot(Me/Mp)) / (h / Me*v) = d*(sqroot(Me/Mp))
d' = 2.7*(sqroot((1.6*10^-27)/(9.1*10^-31)) = 113.214 mm
c) slit seperation =a
d = lambda*D/a
2.7*10^-3 = 3.640*10^-10*2/a
a = 1.348*10^-7m

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