## Sapling hw #7

Natalie 2F
Posts: 40
Joined: Wed Sep 30, 2020 10:11 pm

### Sapling hw #7

Hi! I had a question of one of the problems from sapling which is, suppose the typical work function of the metal is roughly 4.170×10−19 J.
Calculate the maximum wavelength in angstroms of the radiation that will eject electrons from the metal. What would be the first step to solve this? I tried looking at the equations we know to figure out what we need to find but I kept getting stuck.

Connie Liang 3L
Posts: 33
Joined: Wed Sep 30, 2020 10:00 pm
Been upvoted: 1 time

### Re: Sapling hw #7

Hi Natalie! I had some trouble with this one too.

You want to know the wavelength of light that will be JUST enough to remove electrons from the metal with a work function of 3.21 x 10^-19 J.

This means 3.21 x 10^-19 J is exactly equal to the energy of one photon because one photon is equal to JUST the amount of energy it takes to remove an electron. There is no excess kinetic energy because you know the light is of the LEAST amount of energy.

Therefore, you have 2 equations. E = hf for energy of one photon and Energy of photon - energy to remove electron = excess kinetic energy. By setting up your work as hf = the energy to remove one electron (seeing that excess kinetic energy is 0) you can find the max wavelength.

Hope that helps!

rhettfarmer-1l
Posts: 41
Joined: Wed Sep 30, 2020 9:59 pm

### Re: Sapling hw #7

Im going to walk you threw this conceptually. By the equation goes as E(electron)=E(photon)-work function. We are given the work function and they want to know the wave length. The longest wave length is going to be the least amount of energy need to emit this electron meaning that Kintic energy has to be 0. Hence the work function has to equal the energy of the photon or hv. Therefore we can write it as hc/lamanda equal the work function given. Hence you will get the longest wave length and then convert from Meters to Angstrom by dividing by 10^-10 welcome!.