Checking answer for velocity of photoelectron

[eldred]

Homework Statement

Find the speed, v, which an electron, emitted with kinetic energy W = 0.200 eV, arrives at the anode when the accelerating voltage between the anode and the cathode is U = 2.00 V.

b) The value I of the photoelectric current

Homework Equations

From what i gather,
E_total = E_kinetic + work function = qV + work function

The Attempt at a Solution

Im not sure if the first question is poorly worded or I'm just misunderstanding it, but I assume by 'kinetic energy W = 0.200' they mean the work function. I solved it like so

E_total = E_kinetic + work function = qV + work function

E_TOT = ((1.6x10^-19)x(2) + (0.2 x 1.6x10^-19)
= 3.52x10^-19 joules
3.52x10^-19 = 1/2 mv², solving for v i get 879077m/s.

Does the electron actually move 879 KILOMETERS a second? Wow. Is this right? Is it true only because it travels a very short distance in total, over a short time? I'm having a hard time conceptualizing this.

Now for b
I'm really not sure how to approach this, I know current I = voltage / resistance, but it's not applicable in this situation,

from the last question we know that the voltage is (3.52x10^-19)/(1.6x10^-19) which seems to be 2.2 volts? But where do I go from here?If possible I need urgent clarification, I need to go sleep soon. Thanks for any help!

 

[Aceix]

Homework Statement

Find the speed, v, which an electron, emitted with kinetic energy W = 0.200 eV, arrives at the anode when the accelerating voltage between the anode and the cathode is U = 2.00 V.

b) The value I of the photoelectric current

Homework Equations

From what i gather,
E_total = E_kinetic + work function = qV + work function

The Attempt at a Solution

Im not sure if the first question is poorly worded or I'm just misunderstanding it, but I assume by 'kinetic energy W = 0.200' they mean the work function. I solved it like so

E_total = E_kinetic + work function = qV + work function

E_TOT = ((1.6x10^-19)x(2) + (0.2 x 1.6x10^-19)
= 3.52x10^-19 joules
3.52x10^-19 = 1/2 mv², solving for v i get 879077m/s.

Does the electron actually move 879 KILOMETERS a second? Wow. Is this right? Is it true only because it travels a very short distance in total, over a short time? I'm having a hard time conceptualizing this.

Now for b
I'm really not sure how to approach this, I know current I = voltage / resistance, but it's not applicable in this situation,

from the last question we know that the voltage is (3.52x10^-19)/(1.6x10^-19) which seems to be 2.2 volts? But where do I go from here?If possible I need urgent clarification, I need to go sleep soon. Thanks for any help!

Since it was said that the kinetic energy is 0.2eV, you will not need the work functiin of the metal.

 

[gneill]

Homework Statement

Find the speed, v, which an electron, emitted with kinetic energy W = 0.200 eV, arrives at the anode when the accelerating voltage between the anode and the cathode is U = 2.00 V.

b) The value I of the photoelectric current

Homework Equations

From what i gather,
E_total = E_kinetic + work function = qV + work function

The Attempt at a Solution

Im not sure if the first question is poorly worded or I'm just misunderstanding it, but I assume by 'kinetic energy W = 0.200' they mean the work function. I solved it like so

E_total = E_kinetic + work function = qV + work function

E_TOT = ((1.6x10^-19)x(2) + (0.2 x 1.6x10^-19)
= 3.52x10^-19 joules
3.52x10^-19 = 1/2 mv², solving for v i get 879077m/s.

I think they mean that the electron that escapes the surface is left with a kinetic energy W = 0.200 eV. Your calculation treated it this way, so the result looks fine.

Does the electron actually move 879 KILOMETERS a second? Wow. Is this right? Is it true only because it travels a very short distance in total, over a short time? I'm having a hard time conceptualizing this.

Yup. Subatomic particles can move really fast with small amounts of energy. They tend to run into obstacles pretty quickly when they're not in a vacuum.

Now for b
I'm really not sure how to approach this, I know current I = voltage / resistance, but it's not applicable in this situation,
from the last question we know that the voltage is (3.52x10^-19)/(1.6x10^-19) which seems to be 2.2 volts? But where do I go from here?

Current is also the rate of movement of charge: An ampere is a Coulomb per second.

Is there information missing in the problem statement that would allow you to quantify the rate of electron emission?

 

[eldred]

Ah yes! The question I've posted is a sub-question to a larger problem. I was also given this info:
The power absorbed by the photo-cathode is P_c = 0.540 W, but only 0.2% of this incident energy is effective in producing photoelectrons.
I have no clue how i didn't realize the rate of emission is right there.
Current I = power W / voltage V
So would this calculation be correct? I = (0.540 x 0.2) / 2.2 = 0.049 amps?

A big thank you for helping me with this!
:)

 

[gneill]

I would not look to the voltage as a contributor to the current. It doesn't cause the emission of the electrons from the surface of the cathode, it merely accelerates the ones that are emitted.

Instead, consider the rate at which the absorbed energy is being converted to emitted electrons (you know the energy they are each given).