A student studying the photoelectric effect for two different metals records the following information. (i) the...
A student studying the photoelectric effect for two different metals records the following information.
(i) the stopping potential for photoelectrons released from
metal 1 is 1.24 V larger than that for metal 2
(ii) the threshold frequency for metal 1 is 50.0% smaller than that
for metal 2.
Determine the work function for each metal.
ϕ1 =
ϕ2 =
Homework Answers
here V01 = -1.24 V02
V01 = -0.5 v02
eV0 = hf - hvo
hvp = evo1 + hvo1 for metal 1
hvo = evo2 + hv02 for metal 2
from frist two equations
evo1 + hv01 = evo2 + hvo2
e(1.24 v02) + h(0.5 v02) = ev02 + hv02
e(0.24 vo2) + h(0.5 vo2)
hv02 = 24/50 ev02
phi = hv0 = 24/50 ev02
phi1 = 0.50 * 24/50 = 0.24 * 1.6 e -19 V02 = 0.384 Vo2
phi 2 = phi1/0.5 = 0.768 Vo2
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