The electrons in a cathoderay tube are accelerated through a

The electrons in a cathode-ray tube are accelerated through a 250 V potential difference and then shot through a 38-nm-diameter circular aperture. What is the diameter of the bright spot on an electron detector 1.5 m behind the aperture?

Solution

E = qV = e*250 V

wavelength = hC/E = (6.62*10^-34*3*10^8)/(250*1.6*10^-19) = 4.96*10^-9 m

w = 2.44*wavelength*L/D

=>w = (2.44*4.96*10^-9*1.5)/(38*10^-9) = 0.4777 m