# please help whether to do these kind of questions by analysing graphs or other way: i am getting confused the freq and intensity of a light source are both doubled. (1)saturation photocurrent remains almost same (2)max k.e. of photoelectrons is doubled a.both true b.1 true 2 false c.1 false 2 true d.both false please reply soon!! aayush

please help whether to do these kind of questions by analysing graphs or other way:

i am getting confused

the freq and intensity of a light source are both doubled.

(1)saturation photocurrent remains almost same

(2)max k.e. of photoelectrons is doubled

a.both true

b.1 true 2 false

c.1 false 2 true

d.both false

please reply soon!!

aayush

## 1 Answers

7 years ago

Now we have to stick on strictly with the fact that one to one correspondence exists between photon fallen and electron ejected. No two photons would eject single electron and not a single photon could eject two electrons.

Now by the conservation of energy principle, the energy content of the photon is used it two ways. One to do work against to take off the electron right from the surface. This is defined as the work function W. The other part of energy is handed over to that ejected electron in the form of kinetic energy. 1/2 m v^2.

Now let us understand about the way in which we define electric current. Current is defined as the charge flown per sec. As the number of electrons increase then charge also increases. If the intensity ie photons increases then more electrons and hence increase in photo electric current. This is well understood by every one of us.

Now think about the velocity. If higher the velocity, then in one second more electrons would get collected by the anode and so electric current increases. Also we have to stick on to the point that not all electrons coming out due to photo electric emission would have the same velocity. There will be a maximum no doubt which is got by Einstein's photo electric equation hv = hvo + 1/2 m v^2max

As this is the situation, when you increase the frequency then the probability of electrons arriving at anode will be more in one second and hence the current increases.

So it is true that as the frequency increases there will an increase in the photo electric current. It is purely due to the chance.

Of course we cannot say surely that as we double the frequency of radiation, then the photo electric current is also doubled. Because we don't have such a concrete equation to have a relation between frequency and current.

In this case both the statement s are false.

The best way to solve these types of problems is to use your analytical skills rather than rely on graphs.