Question

A beam of electrons with kinetic energy of 1 keV passes through a slit of width a = 1 micro-m. The electron beam is then detected on a phosphorescent scree loacted at a distance D = 1 m from the slit. What is the width W of the "image" of the electron beam?

Answer 1

Answer 2

KE = 1keV = hc/L

so wavelenmgth L = 6.626 e-34*3e8/(1.6e-16)

L = 1.24 nm

so

In interfreence or diffraction pattern

the needed equation is Y = mLR/d---------------1

and d sin theta = mL--------------------2

where L = wavelgnth

m = order = 1,2,3,4, ......... for brigth bands

m = 1.5, 2.5, 3.5, 4.5, ......for dark bands

R is the distance from slit to screen

Y = disatnce from central spot to nth order fringe or fringe width

so

W = 2Y = 2 * 1.24 nm * 1/(1um)

W = 2.48 mm