Achieve # 6 , Weeks 2-4

[506002081]

Hello,

How do you find the work function?? I have tried using the equation work function= h*vphoton - kinetic E electron, but the answer I get doesn't work. Please help!

Thank you

[Hao Tam Tran 3H]

Hi!

When using this equation, you have to keep in mind that the "work function" is basically the energy threshold. Only when the photon exceeds this amount can it eject an electron, and the extra energy after it will be put into the kinetic energy of the electron. Thus, the way you're using the equation should be correct (work function = hv_photon - kinetic energy should give you the amount of energy of the photon that went into ejecting the electron in the first place), so you're going to have to troubleshoot. Maybe you used the wrong units. Remember that a joule is a kg•m²/s², so ensure that your units are in these. If the energy is given in electron volts, convert it joules. If the problem states that there was no kinetic energy, that means that the energy of the photon is exactly at this threshold energy so the work function would just be equal to hv in this situation. Just try to find any way you may have made a mistake, because you are using the equation correctly! Hope that helps!

[irisgu]

Hi, for this question, I began by calculating the energy of a photon used to eject an electron. E=hvphoto and Φ=hvphoton−KEelectron. The work function describes the minimum energy required to eject an electron (aka the threshold energy). Since Φ describes the energy per photon, each photon with this minimum amount of energy will free one electron.

For the second part of the problem, recall that the maximum number of electrons will be ejected when KE=0 and the energy of each photon = work function.

Hope this helps!

[506002081]

Okay that helped! For part b, do we divide the number we got in part A by the energy of the burst of photons?

[Baza AlAbdullah 2F]

For part b, you will divide the total energy the question provides by the answer you got from part A.

Hope that helps!