OCR Physics A G482, Electrons, Waves and Photons, 25th May 2012

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Hi guys, anyone know why the photoelectric current in a photocell circuit is proportional to intensity of the incident radiation?

Cheers :smile:

Reply 201

jake9382

guys is this a am or pm exam ????????

Reply 202

Dale12

Original post by jake9382

guys is this a am or pm exam ????????

Reply 203

Toshiya

Original post by jake9382

guys is this a am or pm exam ????????

Afternoon.

Reply 204

Dale12

Original post by Zobinho

Hi guys, anyone know why the photoelectric current in a photocell circuit is proportional to intensity of the incident radiation?

Cheers :smile:

what section of the specification is that in???

Reply 205

Zobinho

Original post by Dale12

what section of the specification is that in???

2.5.2 The Photoelectric effect
http://www.ocr.org.uk/download/kd/ocr_9587_kd_gce_spec.pdf
scroll down page 27

Reply 206

sweetascandy

Original post by Zobinho

Hi guys, anyone know why the photoelectric current in a photocell circuit is proportional to intensity of the incident radiation?

Cheers :smile:

I don't think we have to know that for this exam; it's not on the spec.

EDIT: Okay, maybe it is :colondollar:

(edited 12 years ago)

Reply 207

Pangol

Original post by Zobinho

Hi guys, anyone know why the photoelectric current in a photocell circuit is proportional to intensity of the incident radiation?

Cheers :smile:

Here's something about this from earlier today.

Reply 208

Dale12

Original post by Zobinho

2.5.2 The Photoelectric effect
http://www.ocr.org.uk/download/kd/ocr_9587_kd_gce_spec.pdf
scroll down page 27

ah, imagine a photocell as a cell that needs em radiation to shine on it to release electrons.. E=hf...
the maximum kinectic energy of the electrons is independent of intensity because only 1 photon will interact with 1 electron... however, increasing intensity increases the number of photons that hit the photocell per second... if more photons are interacting with more electrons per second, then more electrons will be released per second and allowed to flow around the circuit... more charge carriers per second results in an increase in current... hope that helps

Reply 209

sweetascandy

Original post by Dale12

Omg, that makes SO much sense! I would +rep you but "You have reached the limit of how many posts you can rate today!" :colondollar:

Reply 210

Dale12

Original post by sweetascandy

Omg, that makes SO much sense! I would +rep you but "You have reached the limit of how many posts you can rate today!" :colondollar:

im just glad it helped you :smile:

in fairness i completely forgot about that, had to qucikly read through my notes from a couple days ago :biggrin:

Reply 211

JoshC.

Quick question; something which the (useless) textbook hasn't helped me with.

With Waves, what is the difference between Amplitude and Displacement? Are they basically the same numerical value, except Amplitude is always positive whilst Displacement can be positive or negative, depending on whether it is above or below the line?

Cheers :smile:

Reply 212

sweetascandy

Original post by JoshC.

Amplitude = the maximum displacement from the equilibrium position.
Displacement = the displace of any point on a wave from its equilibrium position.

So amplitude is the peak, whereas displacement is just simply the distance of any point of the wave (e.g. it could be halfway between equilibrium position and amplitude) from its equilibrium position.

Hope that helps :smile:

(edited 12 years ago)

Reply 213

Dale12

Original post by JoshC.

my textbook says AMPLITUDE = the maximum displacement of a particle from its equilibrium position... so for instance, if you think of a sin wave.. then its amplitude is at 90 degrees whilst at say 45 degrees its displacement would be 0.707... if that makes sense?

Reply 214

JoshC.

Original post by Dale12

Thanks to you both, but that Maths-y way of describing it really has helped.

Cheers! :biggrin:

Reply 215

Wilko94

Can someone explain to me the relationships between current, voltage, and resistance.

eg.

- If you change the voltage supplied, how does the resistance and current change?
- If you increase the resistance, does the current decrease but the voltage stay the same?

Struggling to get my head round all this!

Reply 216

sweetascandy

Original post by Wilko94

I just learnt this today.
Think of voltage as something which helps to push the current round a circuit.
When there's a high resistance and low current, then the voltage needs to increase so that it can push the current around faster.
On the other hand, when there's a low resistance but high current, the voltage doesn't need to push the current round since it's already high; hence the voltage decreases.

Btw, I don't think this is the actual scientific way of explaining it; it's just to help you understand it. :smile:

Reply 217

sweetascandy

Can someone please explain the experiments for stationary waves?
- Like how stationary waves are formed,
- And how to demonstrate them using microwaves, stretched strings and air columns. (+rep) Thankyou.

Reply 218

Dale12

Original post by Wilko94

for a component, resistance is constant so increasing the voltage means the current increases

however if you increase the resistance the current decreases but the p.d stays the same

V=IR

Reply 219

Wilko94

So say there is a parallel circuit with a resistor on each, 1 thats 1k Ohms, another 5k Ohms.

The 5k Ohm resistor will have a smaller current.

Where as the 1k Ohm resistor will have a greater current.

And if the resistors are in series, then then;

The 5k Ohm will have a larger share of the potential difference?