A group of students recently asked mi tis question... how will the frequency affect the photoelectric current?
The simple answer is frequency will NOT affect the photoelectric current, and the standard response is this: frequency will only affect e energy of individual photon (since E=hf), but the number of photons that will incident on the metal surface per unit time will not change, hence the number of photoelectrons emitted will not be affected. This also means that the charge arriving at the collector's plate per unit charge will not change, implying the same current.
Some students will then argue, when the energy of the photon is higher, wouldn't the max KE be higher, hence the velocity of the photoelectrons reaching the collector's plate be higher, and thus higher current?
Well, it is perfectly correct to say that with higher photon energy, the max KE is higher and hence the velocity will also be higher. However, this does not mean that the current is higher, as the rate that the photoelectrons reaching the collector's plate, and hence the quantity of charge, will not be affected. Remember I = ne/t (e is a constant, n/t is the number of photoelectrons reaching the collector's plate per unit time)
I gave this analogy when I was explaining to my students... with slight modification here :)
If all e students in e audi want to go to the toilet, but I only allow one student to go every min. naturally the first couple of stds dun quite feel e urgency n take their own sweet time to go to the toilet, but towards the end the students would rush to the toilet! Do I have more students going to the toilet per min? Of cos not, 'cause i m still controlling the rate of students leaving! In photoelectric effect, this rate is proportional to the quantity known as intensity, and the energy of each student is like the energy of the photons, each of which is proportional to the frequency!
Note: Intensity is actually equal to the energy incident on the metal surface per unit time per unit surface area, which is also equal to (N/t)hf/S, where S is the surface area of the metal. However, when we say thatthe intensity is changed, by default it is assumed that the frequency remains the same (freq characterises the type of EM waves, ie the 'colour'), and hence by changing the intensity would mean changing the rate of incident of photons on the metal.
Subscribe to:
Post Comments (Atom)
1 comment:
Hiya ... Maybe I can share my thoughts on this since I was initally a bit the confused by the issue at first until I thought through your analogy.
I guess your students (and myself included - but haha, i was also your student) were thinking of 'I=Nqva' where big N is the number of electrons per unit volume of space and v is the velocity of the electrons which in the A-level syllabus and thought that I would be proportional to v.
Initially I thought that N would be a constant independent of v but actually it's actually inversely propotiona to v for a source putting out a contant number of photoelectrons per unit time and the two vs cancel off each other so there's no actual v dependence in the photocurrent generated in the end.
Post a Comment