## Photoelectric Effect Module Question 28

Jason Nguyen_1B
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Joined: Wed Sep 30, 2020 9:34 pm

### Photoelectric Effect Module Question 28

Light hits a sodium metal surface and the velocity of the ejected electron is 6.61 x 105 m.s-1. The work function for sodium is 150.6 kJ.mol-1.
What is the kinetic energy of the ejected electron?

For this question I'm a bit confused. I know that I have to use the equation 1/2mv^2 to calculate the kinetic energy of the ejected electron. Would m be the mass of the electron? Also when I searched up the mass of an electron, the units were in kg. When I calculate the kinetic energy, would I have to convert kg into g?

Earl Garrovillo 2L
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Joined: Wed Sep 30, 2020 9:55 pm

### Re: Photoelectric Effect Module Question 28

Hi! You're right that m would be the mass of the electron but keep the units as kilograms (9.109 × 10^-31 kg). Not only are kilograms the SI unit for mass but since energy in Joules is equivalent to $\frac{kg*m^2}{s^2}$ , keeping the mass in kilograms is necessary to get the correct answer.

Thomas Vu 1A
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### Re: Photoelectric Effect Module Question 28

So, just to make sure, is the work function even necessary for this problem? In other words, is there any other information we get or can interpret from being given the work function value? thanks

IsaacLaw1E
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### Re: Photoelectric Effect Module Question 28

The work function would be used to get the energy of the photon that hit the metal, but that's not needed in the question, so the work function isn't necessary here. It's probably for another part of the question that's somewhere else.

Earl Garrovillo 2L
Posts: 91
Joined: Wed Sep 30, 2020 9:55 pm

### Re: Photoelectric Effect Module Question 28

Right, for this specific part of the problem, the work function doesn't have any use. If I remember correctly, the work function was used in a different problem with the same context.

But since you asked what else we can derive from the work function (aka threshold energy or energy to remove electron):
1. If given the energy of the photon that ejected the electron and work function, we could solve for the kinetic energy using $E_{photon}-\phi =KE_{electron}$
2. Vice versa, if given the kinetic energy and work function we can solve for the energy of the photon using the same equation above
3. Similarly, to eject an electron, the energy of a photon has to be equal to or greater than the work function. So sometimes, the work function is equivalent to the energy of the photon if specified in a problem.

Posts: 88
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### Re: Photoelectric Effect Module Question 28

So in order to answer this question you had to look up the mass of an electron of sodium? I was stuck for so long because I did not think that the question gave us enough info.

Posts: 88
Joined: Wed Sep 30, 2020 9:38 pm

### Re: Photoelectric Effect Module Question 28

OH this is a response to my last question and it might be silly but all electrons have the same mass correct? It doesn't differ based on the element?

rhettfarmer-3H
Posts: 96
Joined: Wed Sep 30, 2020 9:59 pm

### Re: Photoelectric Effect Module Question 28

Joules actually has Kg in it so we dont need to convert it. And Kg is the si unit of mass so we are good there. And yes the work function is not need because we are talking about Kinetic energy of the electron which only needs the mass of an electron and the velocity of it which were both given to us. Therefore, we continue to use Ek=1/2mv^2. shown on the work.
Attachments
IMG_2781.pdf

Posts: 62
Joined: Wed Sep 30, 2020 9:47 pm

### Re: Photoelectric Effect Module Question 28

Jaden Haskins 2F wrote:OH this is a response to my last question and it might be silly but all electrons have the same mass correct? It doesn't differ based on the element?

All electrons share the same mass, so it will always be 9.10956 x 10^-31 kg.

Lily Carlson 1K
Posts: 11
Joined: Wed Sep 30, 2020 10:10 pm

### Re: Photoelectric Effect Module Question 28

Yes; since the mass of an electron is always the same, you just need the velocity to calculate the kinetic energy. The work function information is completely unrelated, so you shouldn't be afraid of not using information in problems if it isn't relevant!