Question

17. Two light sources are used in a photoelectric experiment to determine the work function for a particular metal surface. When green light from a mercury lamp (l 5 546.1 nm) is used, a stopping potential of 0.376 V reduces the photocurrent to zero. (a) Based on this measurement, what is the work function for this metal? (b) What stopping potential
would be observed when using the yellow light from a helium discharge tube (l 5 587.5 nm)?

Answer 1

@douglaswinslowcooper
@wolfe8
do either of you know why V and eV can be used interchangeably in this problem?

Answer 2

eV is energy=work. V is potential. Perhaps the e is assumed with the latter.

Answer 3

@LastDayWork

Answer 4

@JoannaBlackwelder

Answer 5

Find the energy of a photon for a green light & then find the maximum (possible) kinetic energy of the emitted electron (in terms of work function).

Next use the fact that
e * stopping potential = max Kinetic Energy

Answer 6

I found the answers, but i don't get why i the math works.
for part a) 2.27 eV - 0.376V = 1.89eV? the units don't match but the subtraction works?

Answer 7

it's illogical

Answer 8

eV can be considered as the kinetic energy of an electron when it passes through a potential of V

"...but the subtraction works?"
Can you explain what does that mean ??

Answer 9

Do you mean to say that 1.89eV is somehow the correct answer ??

Well that's because you are calculating energy in terms of eV.
1 eV = e Joules
where e is the charge of an electron.

So an electron passing through 1 V of potential has an energy of 1 eV = e Joules
Hence, (technically) the equation is -
2.27 eV - 0.376 eV = 1.89eV

Answer 10

@roadjester
Does that makes sense ??

Answer 11

Sorry, I've been afk. Yeah, that makes sense now. And you did interpret my question correctly. I didn't get why you could subtract a value of volts with a value of electron-volts.
Thanks a lot!

Answer 12

You're Welcome :)