NS FL 3 CP Q9

Post Reply
estherk2020
Posts: 1
Joined: Mon Aug 19, 2019 10:50 pm

NS FL 3 CP Q9

Post by estherk2020 » Mon Aug 19, 2019 10:56 pm

The question asks what gamma rays are capable of. My understanding is that gamma rays (synonymous to gamma decay?) can only emit high-energy photons from proton but cannot emit proton, neutron nor electron. The correct answer to this question, however, is that gamma rays are indeed able to create free radicals and eject electrons... Help!
NS_Tutor_Mathias
Posts: 257
Joined: Sat Mar 30, 2019 8:39 pm

Re: NS FL 3 CP Q9

Post by NS_Tutor_Mathias » Tue Aug 20, 2019 5:48 am

Gamma rays are emitted by, but are not synonymous with, gamma decay, and gamma rays are just a kind of electromagnetic radiation (which is all functionally just one phenomenon, really). And you are right that all EM radiation is mass-less. However, EM waves transmit energy, and energy is the ability to do work - and this is the basis of the photoelectric effect, the phenomenon that absorbed EM radiation can eject electrons (by doing more work than it takes to escape the highest energy level, a quantity called the work function [phi]).

You know all this intuitively already: Incident EM radiation on any surface in your macroscopic world heats it up (leave stuff out in the sun and it gets warm). As we get to higher frequency EM radiation, it starts being able to excite electrons rather than just make stuff warm - this is the functional principle behind UV-induced DNA damage. Gamma radiation, as in this case, is just very high frequency EM radiation and works exactly the same way. But by virtue of being very high frequency, it manages to excite even fairly tightly bound electrons and often do more than excite them and actually eject them from their molecular orbitals (which will generate a radical, by definition).

There are a few things to keep clear here too:
1. Photons and EM waves refer to aspects of the same phenomenon, and usually if we're modeling a singular event of electron excitation, we're going to be talking about photons (the particle model)
2. Frequency, wavelength and energy per photon are linked by E_photon = hf and c = lambda * f (where lambda is the wavelength, f is the frequency, c is the speed of light and h Planck's unreduced constant)
3. Keep in mind that intensity, number of photons per area per unit time, is a very distinct phenomenon from energy per photon. Whether EM radiation is able to excite an electron is independent of intensity and depends solely on the energy per photon (see above, point 2)

And to review the game with the work function real quick. The energy of an ejected electron can be predicted if we know the energy of the incident photon (which we will know if we know it's wavelength or frequency): E_photon_incident = KE_electron + phi
This directly implies that if E_photon_incident (energy of the photon being absorbed) is less than phi (the work function), then KE_electron = 0 (or negative, which is an impossibility). In other words, in those cases no electron is ejected.
Post Reply