Doubt from photoelectric effect

@Rudra_d @manan_uppadhyay @Binod unno Guys I have faced a problem since a long while. In almost every book, it is stated that intensity is independent of frequency I.e. intensity expressed by the number of particles incident per unit area per unit time. But on the other hand, there is another formula that crops up for intensity which is Nhv/At, or in other words the energy incident per unit area per unit time. Occasionally this alternate convention works fine, but sometimes the other one seems to yield the right answer. Can someone explain when to take which convention?

2 Likes

I = E/At
E = nh \nu

I = n h \nu \over At

Here n and \nu can be varied to change intensity. It is mostly mentioned how the intensity is changed.


If we are changing number of particle (with fixed freq.) then also we’re changing energy. If we keep number of particles fixed and change freq., then also we’re changing energy and hence intensity.

3 Likes

I understand this bro but lemme show you directly how they confound us. Finding a handful of examples gimme a sec

1 Like

A option given correct for that vedantu multiple choice question

As you can see, if image 1 were to hold true, then A option would be false. Right?
@Binod

2 Likes

If frequency of light is halved, then to keep intensity same, we have to double the number of particles/photons. So, saturation current may double.

1 Like

Yeah exactly, actually let me clear something out of the way, the vedantu question screenshot I sent was from someone else. My own answer for that question today was ABD, which I guess would have been the same as yours (and is also the one given in the key). But the confusion arises when you compare them with the first image, doesn’t it?

The first image follows a different convention/definition of intensity. It defines intensity as N/At instead of Nhv/At

I hope you see the contradiction now? Or maybe I am seeing ghosts as usual

1 Like

Nah, I did ACD. I realised later that I did wrong.

2 Likes

This is counter-intuitive. There can’t be two diff. definitions of intensity for same purpose. Let me revisit the pic you shared.

2 Likes

We’ve been told N = IA/hv
where N = no. of electrons emitted per second.

3 Likes

Yes bro. Exactly. But image 1 is endorsed by a number of sources, including stackexchange and all the books my friends looked into.

2 Likes

par waham voltage bhi increase karraha hena graph mei

2 Likes

Let’s just talk about saturation current for a moment, which is independent of voltage.

1 Like

mei first image ka bolrahahun bhaai waham i think voltage increase karrahahae ek energy derahahe lekin 2nd question aisa nhi

1 Like

I_s = Ne
(I_s = Saturation current)

1 Like

Bro, the intensity which is given by I = N h \nu \over At is correct. We both agree that intensity can be governed by two factors (i) no. of photons and (ii) frequency.

In almost all cases in which intensity of light is varied, one of these factor is fixed, while the other is varied.

If we fix the freq. and vary the no. of photons, so the saturation current will definetly be varied as each photon is capable to eject an electron. More photons, more electrons ejected and hence more saturation current.

If we fix the no. of photons and vary the frequency, then since no. of photons are fixed, so, no. of ejected electrons will be fixed, so saturation current will be fixed and stopping potential will vary with varying frequency.


But if we decide to fix the intensity, then if we make any change in freq. and we’ll also have to make some changes in no. of photons also to keep intensity same. So, if we half the freq. then we have to double the number of photons, and hence both stopping potential and saturation current will change.

From, I = N h \nu \over At … if I = const. then N$\nu$ const. If N is changed then \nu is changed by itself.


I tried to put forward what I know, and I hope something in this will help.

4 Likes

I_s is directly proportional to no. of ejected electrons = no. of photons striked.= N

Ye hi kehna chah rahe ho na bro?

1 Like

I completely agree with what you said. I cannot remember a time when it didn’t work for me. But I have had trouble in the past explaining this to people who use picture 1 as counterevidence. Like my friend who marked BCD in the vedantu question image

2 Likes

Wait I will share link.

3 Likes

I do not completely agree with what the picture says TBH.

@Vibhu_Kumar @Aastik_Guru @Cooljet123 @AshWin @Seshank @Sanjay_S @everyone please look into this.

3 Likes

At the end of the day, I can only hope my doubt isn’t absolutely dumb (:upside_down_face:

1 Like