Pages: 1 2

 

2010-05-05 17:41:58
Ryan
Member
+1,230|7248|Alberta, Canada

Ya ya, I'll delete it once it's answered. This needs attention, attention that it won't get in some homework thread.

In a photoelectric experiment, a student found that when a certain photoelectric surface was illuminated with monochromatic light with a frequency of 6.00x10^14 Hz, the stopping voltage was 1.40V, and when she illuminated it with light of frequency 5.20x10^14 Hz, the stopping voltage was 1.10V. From this data, find:
A) Planck's Constant
B) Threshold Frequency
C) Work Function of the surface


My teacher said to start with C first, then work backwards to get the rest. But I've been staring at it, saying wtf.

Halp plz.
2010-05-05 17:42:42
Gooners
Wiki Contributor
+2,700|7038

fnet = ma
2010-05-05 17:45:10
Ryan
Member
+1,230|7248|Alberta, Canada

Gooners wrote:

fnet = ma
gtfo fo realz
2010-05-05 18:05:07
Flaming_Maniac
prince of insufficient light
+2,490|7112|67.222.138.85
k so let's see if I can avoid fucking this one up

The stopping voltage (fucked up name imo) is just the voltage it would take to negate the kinetic energy completely. energy = charge(voltage) = eV = .5mv2

But, not all the energy imparted from the light goes into kinetic energy. Some of it goes into freeing the electron from the surface in the first place - that is the work factor. Calculate by finding the energy of the photons hitting the surface (E = hc/wavelength) minus the final energy (eV).

The threshold frequency is how much energy a photon must have to free the particle form the surface, but impart zero kinetic energy. Simply set hc/wavelength = work factor, then solve for frequency.

Now since it gave you two data points, you can solve simultaneous equations for h.
2010-05-05 18:10:47
Ryan
Member
+1,230|7248|Alberta, Canada

I know how to solve for all the different things, but why does it give me two different experiments? Average the two? Wtf.
2010-05-05 18:16:22
Flaming_Maniac
prince of insufficient light
+2,490|7112|67.222.138.85
Because if you work it in the order given, you don't know the work factor. You need to solve the equations simultaneously to find h.
2010-05-05 18:18:00
Ryan
Member
+1,230|7248|Alberta, Canada

Flaming_Maniac wrote:

Because if you work it in the order given, you don't know the work factor. You need to solve the equations simultaneously to find h.
How.
2010-05-05 18:20:33
Flaming_Maniac
prince of insufficient light
+2,490|7112|67.222.138.85
?

You don't know h, you don't know E. You have two values for wavelength. You know the value of c. ?
2010-05-05 18:22:33
Ryan
Member
+1,230|7248|Alberta, Canada

I can find E with qV, then find h with the E = hc/wavelength, but that would only be for one of the experiments....

Last edited by Ryan (2010-05-05 18:22:53)

2010-05-05 18:24:42
Flaming_Maniac
prince of insufficient light
+2,490|7112|67.222.138.85
But qV leaves out the work factor. It takes energy to free the electron, and then it has additional kinetic energy.

E = Wf + K = Wf + qV

Again, if you solve it in the order presented, you need both. If you solve it in the way your teacher told you, you don't.
2010-05-05 18:27:21
Ryan
Member
+1,230|7248|Alberta, Canada

Flaming_Maniac wrote:

But qV leaves out the work factor. It takes energy to free the electron, and then it has additional kinetic energy.

E = Wf + K = Wf + qV

Again, if you solve it in the order presented, you need both. If you solve it in the way your teacher told you, you don't.
I'm still lost. I wrote my AP Calclus test today, so my brain is scrambled and trying to regain its composure.

I just wanna know why I have two experiments going and how I'm supposed to solve for one thing (for each point) when I have 2 values for frequency and 2 values for stopping voltage.
2010-05-05 18:29:43
Flaming_Maniac
prince of insufficient light
+2,490|7112|67.222.138.85
Are you going to solve it the what your teacher said or in order?

If you are solving it in order ignore the stopping voltage and try to solve a.
2010-05-05 18:30:35
Ryan
Member
+1,230|7248|Alberta, Canada

Flaming_Maniac wrote:

Are you going to solve it the what your teacher said or in order?

If you are solving it in order ignore the stopping voltage and try to solve a.
She said something about making the work functions equal to each other (I dunno how that's supposed to work).

You want me to solve for h first?
2010-05-05 18:31:39
Flaming_Maniac
prince of insufficient light
+2,490|7112|67.222.138.85
What order do you want to do them in?
2010-05-05 18:32:10
Ryan
Member
+1,230|7248|Alberta, Canada

Flaming_Maniac wrote:

What order do you want to do them in?
I dunno, as long as I answer all the questions. I don't know where to begin.
2010-05-05 18:33:50
Flaming_Maniac
prince of insufficient light
+2,490|7112|67.222.138.85
Start at the bottom then.

The work functions are equal, but you don't have to set them equal.

Do this:

The stopping voltage (fucked up name imo) is just the voltage it would take to negate the kinetic energy completely. energy = charge(voltage) = eV = .5mv2

But, not all the energy imparted from the light goes into kinetic energy. Some of it goes into freeing the electron from the surface in the first place - that is the work factor. Calculate by finding the energy of the photons hitting the surface (E = hc/wavelength) minus the final energy (eV).
2010-05-05 18:36:16
Ryan
Member
+1,230|7248|Alberta, Canada

Flaming_Maniac wrote:

Calculate by finding the energy of the photons hitting the surface (E = hc/wavelength) minus the final energy (eV).
Well I'll use E = hf considering I don't have wavelength.

Do I do it for both frequencies of light? And when I use h, do I use his actual constant, or am I supposed to use the one I "found" (obviously I haven't found it yet)?

Last edited by Ryan (2010-05-05 18:36:28)

2010-05-05 18:51:11
Ryan
Member
+1,230|7248|Alberta, Canada

Flaming? Halp?
2010-05-05 18:52:40
Flaming_Maniac
prince of insufficient light
+2,490|7112|67.222.138.85
You can convert frequency to wavelength.

I guess you could use both of them and not use h. I would just use h.
2010-05-05 18:57:04
Ryan
Member
+1,230|7248|Alberta, Canada

Flaming_Maniac wrote:

You can convert frequency to wavelength.

I guess you could use both of them and not use h. I would just use h.
Well hf is the same as hc/wavelength.

So what, find the kinetic energy of each beam of light? Then what?
2010-05-05 19:07:40
Flaming_Maniac
prince of insufficient light
+2,490|7112|67.222.138.85
It's not really the kinetic energy in the photon...but yes you find the energy in the photon. Then you subtract the final energy in the electron by using qV. What is left is the energy it took to free the electron, or the work factor.
2010-05-05 19:21:16
Ryan
Member
+1,230|7248|Alberta, Canada

Flaming_Maniac wrote:

It's not really the kinetic energy in the photon...but yes you find the energy in the photon. Then you subtract the final energy in the electron by using qV. What is left is the energy it took to free the electron, or the work factor.
Yea, I know that. So I do that for both beams? I'll get two different work functions!
2010-05-05 20:08:19
Flaming_Maniac
prince of insufficient light
+2,490|7112|67.222.138.85
Work function is dependent only on the material.
2010-05-05 20:12:43
Ryan
Member
+1,230|7248|Alberta, Canada

Flaming_Maniac wrote:

Work function is dependent only on the material.
Omg, why are you making this so hard on me.

Give me the first step. Tell me what numbers to use.
2010-05-05 21:45:21
Flaming_Maniac
prince of insufficient light
+2,490|7112|67.222.138.85
Dude I am not making this hard on you. I told you exactly what to do in post #4 with units/equations included. I'm not plugging in the numbers for you.

The work function is a property of the material. It doesn't matter what the intensity of the light is shined on it, it always takes the same amount of work to free the electron. The only difference is how much energy the electron has left when it leaves the material.

 

Pages: 1 2

Board footer

Privacy Policy - © 2025 Jeff Minard