A gap of the width s is irradiated with electrons of the kinetic energy Ek. The

Question

A gap of the width s is irradiated with electrons of the kinetic energy Ek. The intensity of the Electrons I (e) are measured with a detector at a distance D from the gap (see figure). It results a diffraction pattern of the shape. (T(s/A)sin y (b) Preserve the blurring of the position and the impulse of the electrons. Note: Find a relation between Ap and the width of the main maximum). (c) Show that for this experiment the uncertainty Ap.Ax2 h holds. A X screen electron detector 0 Please answer b and din detail. Explain the steps if possible. Thank you!

Explanation / Answer

given electron energy = Ek

now, Ek = 0.5mv^2 ( where m is mass of electron and v is the speed of the electron)

also, from debroglie's waqve equation,. let lambda be the wavelength of the electrons moving at speed v

then lambda = h/mv ( where h is planks constant)

hence

lambda = h/sqroot(2Ekm) = h/p ( where p is the momentum of the electrons)

now, intensity distribution on the screen is given by

I(theta) = sin^2(pi(s/lambda))sin(theta)/(pi(s/lambda)*sin(theta))^2 [ where theta is the angle of of the point of concern on the screen from the slit center and s is the slit width]

let 2*phi be the width of the central maximum

then

I(phi) = 0

0 = sin^2(pi(s/lambda))sin(phi)/(pi(s/lambda)*sin(phi))^2

0 = sin(pi(s/lambda))

hence

pi*s/lambda = n*pi

s/lambda = n

for central maxima, n = 1

s = lambda

when this happens the width of the central maxima is given by

2y = 2*lambda*D/s = w ( width of the main maximum)

2*lambda*D/s = w = 2*h*D/ps

hence

a. p = 2*h*D/sw [ where w is the width of the central maximum]

b. now

p*w = 2*h*D/s ( w is the width of the centrla maxima and p is the mommentum of the electrons)

now, dp*dx = p*w = 2hD/s

now, s is of the order of the wavelength of the electron beam, D is many orders greater

hence 2D/s > 1

hence dp*dx > h/2*pi

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