W = ∆KE = KEf - KEi
So if a certain voltage Vo is applied and the motion of the photoelectron is completely opposed, KEf = 0.
Meaning, W = - KEi. Is it safe to say that in this case, W is negative as the photoelectron is doing work travelling against the electric field set up. Such that W = KEmax, where W = qV?
I ask this because of this question:
One electron ejected from a clean zinc plate by UV light has kinetic energy of 4x1019 J.
What would be the kinetic energy of the electron when it reacher the anode if a retarding voltage of 0.9V was applied between the anode and cathode?
So as W = ∆KE = KEf - KEi, and the work done by the photoelectrons is W = -qV = -1.44x10-19 as the photoelectrons are doing work by opposing the field. which results in:
-1.44x10-19 = KEf - 4x1019
KEf = (4 - 1.44)x10-19
= 2.56 x10-19 J
which is the correct answer.
If you were to take W as positive, then you would end up with a final kinetic energy greater than it's initial which is obviously incorrect.
So my overall question is, is the work done by the photoelectron in this case negative as work is being done "by the body" meaning its kinetic energy is lost by doing this work against moving against the electric field?
Thanks in advance.
So if a certain voltage Vo is applied and the motion of the photoelectron is completely opposed, KEf = 0.
Meaning, W = - KEi. Is it safe to say that in this case, W is negative as the photoelectron is doing work travelling against the electric field set up. Such that W = KEmax, where W = qV?
I ask this because of this question:
One electron ejected from a clean zinc plate by UV light has kinetic energy of 4x1019 J.
What would be the kinetic energy of the electron when it reacher the anode if a retarding voltage of 0.9V was applied between the anode and cathode?
So as W = ∆KE = KEf - KEi, and the work done by the photoelectrons is W = -qV = -1.44x10-19 as the photoelectrons are doing work by opposing the field. which results in:
-1.44x10-19 = KEf - 4x1019
KEf = (4 - 1.44)x10-19
= 2.56 x10-19 J
which is the correct answer.
If you were to take W as positive, then you would end up with a final kinetic energy greater than it's initial which is obviously incorrect.
So my overall question is, is the work done by the photoelectron in this case negative as work is being done "by the body" meaning its kinetic energy is lost by doing this work against moving against the electric field?
Thanks in advance.