How Does Photon Wavelength Affect Electron Ejection from Nickel?

[jackjones12]

Homework Statement

Q. How much energy will it take to remove only one electron on the surface of Nickle? If you hit the surface of Nickel with a photon that has a wavelength of 90 nm, what will the wavelength of that ejected electron be?

Homework Equations

KE= 1/2 MV^2
$$\lambda$$=h/mv
ET=IE+KE

The Attempt at a Solution

I have attempted to change the Et=90nm into velocity and got V=8087.240m/s
Got the IE of Ni which is IE=1.257J
But I am stuck there i have no idea how to get back in the problem to solve for KE then turn that back into nm. Any help will be appreciated. Thnks

 

[JesseC]

erm... well you need to know the 'work function' of Nickel, normally this is measured in electron volts eV. Assuming EI is what you've called the work function, and you've used the unit Joules, that seems wrong. Its far too big so best to double check that.

A physics teacher told me a good analogy for photoelectric effect:

Imagine I give you some money and tell you to get out of the country as far away as possible. Suppose you're on an island, let's say the UK, so you need to get a plane. You need to spend money on the plane ticket obviously, but also you need to spend money getting to the airport! If I don't give you enough money to even make it to the airport, you'll never get out. If I give you a bit more money, maybe you can get to the airport and buy a ticket to France. If I give you loads of money, you can get to the airport and buy a ticket all the way to Japan.

Replace money with energy, UK with metal, yourself for the electron and I'm sure you get the picture... sorry if I'm patronising you.

Some of the photon energy is spent just getting the electron out to the surface of the metal, that's what is known as the work function. The rest of the energy of the photon will go towards the kinetic energy of the electron.

 

[fluidistic]

Homework Statement

Q. How much energy will it take to remove only one electron on the surface of Nickle? If you hit the surface of Nickel with a photon that has a wavelength of 90 nm, what will the wavelength of that ejected electron be?

Homework Equations

KE= 1/2 MV^2
$$\lambda$$=h/mv
ET=IE+KE

The Attempt at a Solution

I have attempted to change the Et=90nm into velocity and got V=8087.240m/s
Got the IE of Ni which is IE=1.257J
But I am stuck there i have no idea how to get back in the problem to solve for KE then turn that back into nm. Any help will be appreciated. Thnks

I agree with JesseC, look for the work function of nickel somewhere on the Internet for example. It's the minimum energy you need to give to an electron on the metal for it to leave it, with no kinetic energy.
When they ask for the wavelength of the electron, I'm guess they mean the De Broglie wavelength, in which case you'd have to apply the formula $$\lambda = \frac{h}{p}$$ where p is the momentum of the electron.

 

[jackjones12]

Thing is we haven't even gotten to work functions so I have to get the answer just using what i have. Ionization Energy of Nickel which is 757kj/mole Then i just factored it down to joules.
757(1000J/1KJ)(1mole/6.022e23)= 1.257J

ET=Total Energy
IE=Ionization Energy
KE= Kinetic Energy

I think I was given the Total Energy in wavelength of 90nm, but i might be reading the problem wrong.

I then turned that into m/s using $$\lambda$$=h/mv
which gave me 8087.240m/s

I thought that i could then just plug in that into KE= 1/2 MV^2
KE=1/2(9.109e-31)(8087.240)^2


But I am not sure if that would even be correct. Even so if it was I am not sure how i would go about turning the answer KE=2.979e-23 Joules back into nm

 

[JesseC]

Thing is we haven't even gotten to work functions so I have to get the answer just using what i have. Ionization Energy of Nickel which is 757kj/mole Then i just factored it down to joules.
757(1000J/1KJ)(1mole/6.022e23)= 1.257J

So let's assume in your case that the ionisation energy of one atom is equal to the work function... then are you telling me:

$$\frac{7.57 \times 10^{6}}{6.02 \times 10^{23}} = 1.257$$

You're missing a very large factor from that answer.

You then want to work out the energy of the photon. Remember this formula:

$$E = h f = \frac{h c }{\lambda}$$

So energy of the photon minus work function should give you the kinetic energy of the electron which you can put into the normal KE formula.