## HW 1B15

$c=\lambda v$

Maeve Miller 1A
Posts: 51
Joined: Fri Aug 09, 2019 12:16 am

### HW 1B15

Hi, guys! Does anyone know why you can't use the c=λv equation to derive the wavelength for part (c)? In the solutions manual, they add together the energy put in (hv) to the energy that is leaving (1/2mv^2) and then use those to find the wavelength, but I'm confused as to why that is a necessary step.

Hannah Lee 2F
Posts: 93
Joined: Thu Jul 11, 2019 12:15 am

### Re: HW 1B15

Hi, guys! Does anyone know why you can't use the c=λv equation to derive the wavelength for part (c)? In the solutions manual, they add together the energy put in (hv) to the energy that is leaving (1/2mv^2) and then use those to find the wavelength, but I'm confused as to why that is a necessary step.

According to the photoelectric effect, an electron is ejected from the metal surface with KE when radiation of high enough frequency (low enough wavelength) is directed at it. This minimum amount of energy required to remove the e- from the metal is called the work function or the threshold energy. We already solved for that work function in part (b), since the question implies that your answer is the minimum amount of energy needed to eject electrons.

If the energy of the photon (hv) >= work function, then energy is ejected with KE of Ek = 1/2mev2, where v = speed of electrons emitted and m = mass of electron. This is expressed in the equation: 1/2mev2 (KE of ejected e-) = hv (E of incoming photon) - work function (E required to eject e-, or threshold E). In this case, we would be solving for hv, or the E of the incoming photon. We don't use the c = λv equation directly because we have to take into account the KE of the ejected electron (which was provided to us in the question). In other words, since the electron is ejected with a certain KE (v = 3.6 x 10^3 km/s), we know that the amount of energy needed to photoeject the electron will be greater than just the bare minimum threshold energy.

So rearranging that equation to solve for the energy of the incoming photon gives us: E = 1/2mev2 + work function. After plugging in the values, you receive the energy of the photon, and you can rearrange c=λv and E = hv to solve for λ via λ = hc / E.