help (1 Viewer)

[chilli 412]

i did this photoelectric effect question a while ago but got a different answer to my teacher and the book it came from

When yellow-green light with a wavelength of 500nm is shone on a metal, the photoelectrons require a stopping voltage of 0.8V. Calculate the work function of the metal in electron-volts.

my method was to just find the work function normally and then convert to electron volts at the end

here is my working:

however the answer is supposedly 1.68 eV
am i right or am i going crazy

 

[jimmysmith560]

Would the following working help?

 

[chilli 412]

Would the following working help?

did you set Kmax = V instead of Kmax = Vq because we would eventually be dividing by q to convert into electron volts? however q is -1.602*10^-19, not 1.602*10^-19

if we assume that after division, a coefficient of -1 still remains as the charge is negative, we end up with the 3.282 eV result. what is the justification for removing 'q' from the equation ?

 

[carrotsss]

i did this photoelectric effect question a while ago but got a different answer to my teacher and the book it came from

When yellow-green light with a wavelength of 500nm is shone on a metal, the photoelectrons require a stopping voltage of 0.8V. Calculate the work function of the metal in electron-volts.

my method was to just find the work function normally and then convert to electron volts at the end

here is my working:
View attachment 38647
however the answer is supposedly 1.68 eV
am i right or am i going crazy

You don’t do the charge as negative for electron volts, keep in mind it’s just a unit so you treat it as such

 

[chilli 412]

You don’t do the charge as negative for electron volts, keep in mind it’s just a unit so you treat it as such

does that mean my method isnt valid? i believed i should be able to work normally with joules and then convert to electron volts once i finished

 

[carrotsss]

does that mean my method isnt valid? i believed i should be able to work normally with joules and then convert to electron volts once i finished

You can still use your method, you just have to treat electron volts and joules as units, so divide by 1.602*10^-19 to get to eV and multiply by 1.602*10^-19 to get back to J, you don’t include the minus because it kinda doesn’t really make sense to do so. If you look in that question, you basically changed it so that the voltage is negative and there are no electrons being released when that’s not the case, which is why you ended up with a much bigger work function than you should’ve.

 

[chilli 412]

You can still use your method, you just have to treat electron volts and joules as units, so divide by 1.602*10^-19 to get to eV and multiply by 1.602*10^-19 to get back to J, you don’t include the minus because it kinda doesn’t really make sense to do so. If you look in that question, you basically changed it so that the voltage is negative and there are no electrons being released when that’s not the case, which is why you ended up with a much bigger work function than you should’ve.

i meant that the quantity Vq wold be negative since charge is negative

i thought since Work = Vq and work is being done against the electrons (the direction of force is opposite to the direction of motion) which means that our work must be negative which implies that Vq is negative, which makes sense since V is positive and q is negative

but tbh i dont really understand work that well, i just thought i had to use q = -1.602*10^-19 since q represents the charge of an electron in this instance and the reference sheet gives the charge of an electron as -1.602*10^-19

 

[carrotsss]

i meant that the quantity Vq wold be negative since charge is negative

i thought since Work = Vq and work is being done against the electrons (the direction of force is opposite to the direction of motion) which means that our work must be negative which implies that Vq is negative, which makes sense since V is positive and q is negative

but tbh i dont really understand work that well, i just thought i had to use q = -1.602*10^-19 since q represents the charge of an electron in this instance and the reference sheet gives the charge of an electron as -1.602*10^-19

The work function is effectively the ionisation energy for each electron, so you can ignore the direction of the work, since the electron uses energy when it is ionised.

 

[chilli 412]

The work function is effectively the ionisation energy for each photon, so you can ignore the direction of the work, since the electron uses energy when it is ionised.

wdym

 

[carrotsss]

wdym

The way the photoelectric effect works is that when a photon has greater energy than an electrons ionisation energy (E=hf) and collides with it, the electron is ionised, and obviously the amount of energy consumed for this to occur (the work) is positive since it’s potential energy from the nucleus is being increased as it goes further away (just like satellites etc). This amount of the energy is the work function, and that’s where the formula K=hf-Φ comes from - any excess energy once the photon has done the work to ionise the electron is transferred into kinetic energy for the electron. Since the ionisation of the electron uses up energy from the photon, logically the remaining kinetic energy will be less than the original energy of the photon.

To explain it with the W=qV formula (I don’t bother using this for photoelectric effect questions, you just divide by the unit for eV on the data sheet usually but I guess it’s useful to have the conceptual understanding), I think the reasoning is that the stopping voltage is the voltage required to counteract the ionisation, so in the formula W=qV the voltage is actually negative. That could be the wrong explanation though, I just came up with that conclusion by combining other physics knowledge

 

[chilli 412]

The way the photoelectric effect works is that when a photon has greater energy than an electrons ionisation energy (E=hf) and collides with it, the electron is ionised, and obviously the amount of energy consumed for this to occur (the work) is positive since it’s potential energy from the nucleus is being increased as it goes further away (just like satellites etc). This amount of the energy is the work function, and that’s where the formula K=hf-Φ comes from - any excess energy once the photon has done the work to ionise the electron is transferred into kinetic energy for the electron. Since the ionisation of the electron uses up energy from the photon, logically the remaining kinetic energy will be less than the original energy of the photon.

To explain it with the W=qV formula (I don’t bother using this for photoelectric effect questions, you just divide by the unit for eV on the data sheet usually but I guess it’s useful to have the conceptual understanding), I think the reasoning is that the stopping voltage is the voltage required to counteract the ionisation, so in the formula W=qV the voltage is actually negative. That could be the wrong explanation though, I just came up with that conclusion by combining other physics knowledge

just out of curiosity how would you have answered the original question if they didnt ask for electron volts and just wanted you to work in joules without making the conversion

 

[carrotsss]

just out of curiosity how would you have answered the original question if they didnt ask for electron volts and just wanted you to work in joules without making the conversion

I’d find the frequency of the light

And then I’d just use there formula from the formula sheet, so

Which you can verify is correct by dividing by 1.602*10^(-19)

 

[chilli 412]

I’d find the frequency of the light

And then I’d just use there formula from the formula sheet, so

Which you can verify is correct by dividing by 1.602*10^(-19)

but how come you used positive 1.602*10^-19 rather than -1.602*10^-19 for the charge of the electron
im sorry for sounding silly but i still dont rlly get the understanding behind it

 

[carrotsss]

but how come you used positive 1.602*10^-19 rather than -1.602*10^-19 for the charge of the electron
im sorry for sounding silly but i still dont rlly get the understanding behind it

From purely an actual understanding perspective, stopping voltage is the work which is required to reduce the kinetic energy of all the electrons to zero (including the most energetic electrons, the ones with Kmax). By its nature, to get from E=a to E=0 for instance (where a is an arbitrary number) would require a work (change in energy) of W=-a. To be pedantic I should really have done a negative symbol in front of qV and then used a negative q value, but they cancel out anyway so it effectively doesn’t matter.