Question

An electric potential is set up across a pair of parallel plates such that the left plate is at -509 V and the right plate is at 0V. An electron is released at the left plate and accelerates towards the right plate. When it reaches the right plate, it passes through a small hole and enters a region where there is a magnetic field of strength 2.690×10-3 T pointing into the paper. The electron travels in a semi-circular orbit and hits a detector a distance, d, from the point where it first entered the region of the magnetic field. See the figure below. MassSpect01.gif What is the kinetic energy of the electron when it reaches the hole in the right plate? Submit Answer Tries 0/10 What is the speed of the electron as it passes through the hole? Submit Answer Tries 0/10 What distance, d, will place the detector at the appropriate position to detect the electron?

Answer 1

(i) Kinetic energy of the electron = KE = eV = 1.6*10^-19 * 509 = 8.14 x 10^-17 J or 509 eV

(ii) v = sqrt (2ev/m)

= sqrt (2* 1.6*10^-19 * 509/9.11*10^-31)

= 13.37 x 10^6 m/s

= 1.337 x 10^7 m/s

(iii) Again, mv^2/2d = qvB

=> 2d = mv/qB

= 9.11*10^-31 * 1.337*10^7/(1.6*10^-19 * 2.69*10^-3)

= (12.18 x 10^-24) / (4.304 x 10^-22)

= 2.83 x 10^-2 m = 2.83 cm.