## HW 1.33b

$\lambda=\frac{h}{p}$

Manya Kidambi 3I
Posts: 31
Joined: Fri Sep 28, 2018 12:24 am

### HW 1.33b

For homework problem 1.33, why is it that E = h$\nu$ is used in part b as opposed to the work function? Isn't the energy required to release the electron the same as the threshold energy, or $\phi$?

mahika_nayak_3L
Posts: 60
Joined: Fri Sep 28, 2018 12:26 am

### Re: HW 1.33b

Basically, because the electrons are simply ejected, it is reasonable to assume that the work function (energy only needed to remove an electron) is equal to the energy of the photon (E photon = hv). By setting these E photon = work function, you can substitute hv and solve from there for the energy of the photon needed to remove the electron.

clamond3F
Posts: 8
Joined: Fri Sep 28, 2018 12:23 am

### Re: HW 1.33b

because you are looking for the exact energy needed to eject the electron and no more, solve the equation
(Energy of photon) - (work function) = (Electron Kinetic Energy)
with kinetic energy = 0 so you're essentially solving for when (Energy of Photon) = (work function). Therefore you can solve the problem by substituting hv for the work function because (Energy of photon = hv).