21if frequency increases , the energy will increase.... but the no. of photons striking per second remains same.... what increases is the kinetic energy of the electrons
Q.It is given in theory books that the magnitude of photoelectric current is independent of the frequency of incident radiation .
My doubt : If frequency is increased , each electron's energy increases which means some of the electrons which were previously unable to reach the other plate [ the more positive plate ] will now have the energy to reach that plate , so eventually Photocurrent WILL INCREASE ??? [7]
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8 Answers
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1ya..exactly that is what i'm saying...when kinetic energy inc , more electrons can reach the opposite plate [ as initially some electrons were not able to reach the other plate to smaller kinetic energy and collisions..]
1I know that :
When intensity inc : num of elec inc and current inc
When freq inc : energy of each elec inc and not inc in their number..but tell abt my post #3 [7]
39How will more of them be able to reach the other end? There could be even more collisions, more slowdown in speed, etcetc...wonder if drift velocity can apply here.
1Initially and finally , there will be same num of collisions rt ?..[why not ? ]
1I dont think the concept of current and drift vel can be actually applied because this is not a kind of restricted region where you can have electrons unlike in a conductor..?
1Anyone there to answer this ??:..
I will give one MORE EXPLANATION .....
MY Explanation : Assume that initially 6 th shell is the valence orbit of a metal atom.I take a source of frequency [ means 'photons; itself ] 'F' .With this value of frequency [and hence a definite amount of energy ] I'm able to take the only 2 electrons in 6s PLUS 1 electron in the 5d [it is the next outermost electron containing orbital ] orbital [ assume that i have ONLY 3 PHOTONS.so 2 photons hit two 6s electrons and one hits the 5d electron. .] out of which the two 6s-electrons go and hit the opp plate with a kinetic energy , thus creating photo-current WHILE THE 5d electron [ having lower energy than 6s electrons according to Aufbau principle ] ends with ZERO KINETIC ENERGY [Even though all the 3 electrons undergo collision b4 coming out , the less energetic 5d electron has more chances of ending up with ZERO KINETIC ENERGY than others rt ? ]
Now , I take a source of frequency '2F'.with this I'm able to make all the three electrons go and hit the plate [ as I am giving more energy to the 5d electron now [than b4 ] and it now has enough kinetic energy to go and hit the plate ! ].so ONE MORE ELECTRON CAN NOW GO AND HIT THE ANODE !.SO PHOTO-CURRENT INCREASES RT ?
39It's as if you're comparing this to kinetic theory...:\
There is no such relation which states that more electrons can necessarily go and strike the positive plate if K.E. increases. That was a postulate of the modern kinetic theory, which placed importance on orientation and kinetic energy of particles(not electrons).
It said that a "probability" factor depended on K.E., as to how many particles could collide and cause a reaction(if they have the correct orientation).
Though it seems like common sense, that if you increase frequency, the increase in K.E. should cause more electrons to strike, that is exactly what has been disproved by the kind and honourable gentlemen who first conducted the photoelectric experiment.
And photoelectric current for various frequencies of IR of a metal having a particular work function, remains the same as long as intensity of the incident beam does not change.
What you say about more electrons going and striking the plate depends on the intensity of IR(incident ray), not on the frequency. Increasing the intensity does increase photoelectric current, increasing frequency does not.
Save yourself jsg. BITS won't ask these.