00:02
So in this problem, we're dealing with the photoelectric effect and must also use kirkoff's rules.
00:08
So for part a, we want vac if r equals 3 .20.
00:14
So i equals epsilon, req, which is 15 .0 volts over 4 .20, and that is 3 .5714 mas.
00:28
So vr is equal to r times i, which is 3 .20 times 3 .5 .714 mas, and that leaves you with 11 .4 vs.
00:49
For part b, we want vac if r equals 334, so vr is equal to 334 times 15 volts over 1 .334k.
01:02
That gives you 3 .76 volts.
01:13
So, ev not equals hc divided by wavelength minus that.
01:19
So if we solve for hc over wavelength minus ev not, we get hc over 140 nanometers, minus e3 .76 volts is equal to 5 .10.
01:41
Evs.
01:44
For part d, we want i, so we need to apply kirkcroft's rules using the currents shown in the figure.
01:51
So i -15 is equal to i plus i, and so a counterclockwise path through the lower part of the circuit gives r -i -minus 1 .0 times i -15 plus 15 volts, and that equals zero.
02:14
So if we solve for capital i, lowercase, i equals 3 .71 mas, and r equals 334...