Question

Help with a physics lab about the Photoelectric Effect

Answer 1

@Data_LG2 this is the website that we are running the lab on:
http://phet.colorado.edu/en/simulation/photoelectric

Answer 2

Questions:

3. 3. Keep battery voltage at 0. Turn light intensity up to 100%. You will be testing sodium first (metals are changeable in the upper right hand box).
4. Adjust wavelength to a value which just allows electrons to leave the surface at a lowest possible speed.
5. Calculate W in electron volts using the following values and formulas:

c = , where c = 2.998E17 nm/sec,  is in nm and  is in s-1
h = 6.626E-34 Js
1 electron volt = 1.60217646 × 10-19 joules
h = Ek + W
***remember—Ek is set at zero

6. Calculate work function (in eV) for all other metals, including the mystery metal.
7. Look up work function values on the internet. Identify the mystery metal, and check the values obtained for the other metals.

Questions:

1. Choose a metal and choose a wavelength that ejects electrons at a reasonable speed. Turn the light intensity up and down. Make note of the number of electrons ejected. Why does the number change?
2. Using this same wavelength and 100% intensity, change the flow of electricity from the battery (at the bottom of the screen), making note of the change in charge at both ends of the electron chamber. Speculate as to why the flow of electrons is changed.

Answer 3

The biggest issue I am having is the formula S:

Answer 4

Just realized the formula didnt paste correctly

Answer 5

\[c=\lambda v\]

Answer 6

\[hv= E _{k}+ W\]

Answer 7

be right back

Answer 8

ok....

I'll try to find out how to calculate eV while you're gone :)

Answer 9

I am back

Answer 10

Sorry I went to take a shower quick

Answer 11

Alright... so I assume that from the simulation, you can only have the values for wavelength (lambda).

\(\sf c=\lambda v, \\ hv= E _{k}+ W \)
we can arrange the the 1st formula in terms of the frequency and sub it into the 2nd formula
\(\sf c=\lambda v, \\ v= \frac{c}{\lambda}\)

insert in 2nd equation and you'll get:
\(\sf W= h(\frac{c}{\lambda})-E_k\)

we can solve for the W first in terms of Joules, then we can convert it to eV later.

So have you gathered the important data (wavelengths with the corresponding number of electrons with each wavelength)?

Answer 12

Could you help judging the wave length. Like for me the slowest it moves without producing more electrons is 538 nm

Answer 13

like anything lower than that the electrons stop :C

Answer 14

i mean higher oops

Answer 15

I'm not sure how to help you with that part because I can't be able to download or run the application..... well, you can try approximating it.

Are you doing a written lab report about this? If yes, then you can probably mention it to under the Discussion part of the report. You can include it as one of the "errors" in this lab.

Answer 16

Hokay so question what does 6.626E-34 is that saying e= 2.71828183

Answer 17

no, upper case E always means \(\sf 10^n\)

Answer 18

so 10^-34?

Answer 19

yes

Answer 20

\[6.626*10^{-34}(\frac{ 2.998 *10^{17} }{ \lambda })-10_{k}\]

Answer 21

Ek = 10?

also, what did you get for the wavelength from the simulation?

Answer 22

So the closest i get is 538

Answer 23

nm

Answer 24

I ws confused s: i dont know what Ek is

Answer 25

oh okay...
the instruction says: "***remember—Ek is set at zero" unless it is changing in the app?
where did you get 10 though? i'm just curious

Answer 26

OH rip

Answer 27

Sorry was a brain forgetful ness moment

Answer 28

haha it's okay :P

Answer 29

because E always means 10 S: i thought

Answer 30

ahh

if it says \(\sf E_k\) this means the kinetic energy.
But if the upper case E is included in some digits, it means \(\sf 10^n\)

Answer 31

Hokay so then

6.626∗10^−34(2.998∗10^17/538)−0 ?

Answer 32

yes, I think so :)

Answer 33

So many numbers T_T

Answer 34

you can do it! :D
i'll grab my calc too

Answer 35

uhm S:

http://www.wolframalpha.com/input/?i=%286.626%E2%88%9710%5E%E2%88%9234%29%282.998%E2%88%9710%5E17%2F538%29

Answer 36

I just use wolfram as a calculator/mathway

Answer 37

cept when i was doing that sin and tan stuff T_T that stuff was annoying

Answer 38

lol alright, that's good then..

now what will be the unit for that?

Answer 39

3.69 x 10^-19 Joules?

Answer 40

yes, now convert that in terms of eV

remember: 1 electron volt = 1.60217646 × 10^-19 joules

Answer 41

Oh lord so much lol

Answer 42

so does this mean 3.69 electron volts I am confused s:

Answer 43

tip: if you are confused with converting stuffs, always look at the units.

the given: 1 electron volt = 1.60217646 × 10^-19 joules
this is eV: J

Now, you have 3.69 x 10^-19 Joules, and you are looking for eV
follow the order of ratio above: eV: J

Now let's do some ratio and proportion:

1 eV : (1.60217646 × 10^-19 joules) = eV : (3.69 x 10^-19 Joules)

in fraction form: