Question

Imagine that you want to run a single-quanton two-slit interference experiment with electrons. If the distance between your electron gun and the electron detectors is about 1 m, your electrons have a kinetic energy of 90 eV, and you want there to be only a 1/10000 chance that an electron is in flight at any instant, what should the current represented by the electron beam be (in femtoamperes)?

Hint 1: Make the approximation that the electron is moving non-relativistically.
Hint 2: Assume that the probability per second of emitting an electron is set at a constant value for all time. What should this probability per second be in order to get the expected 1/10000 chance of an electron being in flight? You can set your electron emission rate equal to this probability per second.

Provided Constants:
e=1.60×10−19 C
1eV=1.602×10−19 J
me=9.11×10−31 kg

ANSWER SHOULD BE IN FEMTOAMPERES

Answer 1

The kinetic energy of the electron beam is, (90ev)(1.6x10-1J/ev) (0.5)(91110k) _1.44x10-17J 4.555x103T Thus, the speed of beam is, v5.6225x10° m/s The time taken to travel is, 5.6225×106 m/s Therefore, the current is, C)! 1.78x10s da dt (1.6×10- 10000)= 8.98×10-17 A(or)0089A