For the question on conservation of energy, what did everyone do, i added the powers, 50.4 i think i got and as p=IV, and current was 4.2, i divided 50.4/4.2, which gave me 12v, the amount which the cell was supplying, is this right?

Posted from TSR Mobile

Reply 1467

Goods

16

Original post by Qari

I forgot to mention keep y the same only talked about x how much do you think i'll get?

It definitely wouldn't cost you more than a mark, so i'd say 2 out of 3?

Reply 1468

swagadon

7

Original post by BayHarborButcher

That's what I put but I'm the only person I know that did :frown:

It may reduce their energy but they're still being emitted surely? (Until it's below the threshold)

ive put the photons will have less energy, and therfore the electrons will have less kinetic energy

Reply 1469

Qari

20

Original post by zedd01

For the question on conservation of energy, what did everyone do, i added the powers, 50.4 i think i got and as p=IV, and current was 4.2, i divided 50.4/4.2, which gave me 12v, the amount which the cell was supplying, is this right?

Posted from TSR Mobile

I did something like that, I added the others and I did 12*4.2

Reply 1470

Zakee

Original post by swagadon

ive put the photons will have less energy, and therfore the electrons will have less kinetic energy

I said the maximum kinetic energy of the photoelectrons decreases, but the number of photoelectrons emitted remains the same (as the intenstity remains the same).

Reply 1471

Qari

20

Original post by swagadon

ive put the photons will have less energy, and therfore the electrons will have less kinetic energy

I've put lower frequency so less will be emitted due to less Ek

Reply 1472

swagadon

7

Original post by Qari

It does the kinetic energy would be lower

doubling the intensity means doubling the number of photons given out per meter squared,so as there are double the amount of photons( the photons frequency has not changed), there will be double the amount of photoelectrons emitted

Reply 1473

123456789012

11

Original post by zedd01

For the question on conservation of energy, what did everyone do, i added the powers, 50.4 i think i got and as p=IV, and current was 4.2, i divided 50.4/4.2, which gave me 12v, the amount which the cell was supplying, is this right?

Posted from TSR Mobile

Yeah, I added the powers too and I got 50.4W. Then, I used P=I^2V and got 50.4 again. This showed that the energy transferred was conserved.

Reply 1474

ac3xx

1

Original post by BayHarborButcher

That's what I put but I'm the only person I know that did :frown:

It may reduce their energy but they're still being emitted surely? (Until it's below the threshold)

The lower the frequency, the lower amount of kinetic energy each electron has until the frequency goes below the threshold frequency, where no electrons are then emitted, but the amount of electrons emitted remains the same. Check question 4a) on June 2009, it's a similar question but the opposite way (asking about an increase).

Reply 1475

Qari

20

Original post by Zakee

I said the maximum kinetic energy of the photoelectrons decreases, but the number of photoelectrons emitted remains the same (as the intenstity remains the same).

No but the ones deeper in the metal might not get emitted due to lower Ek

Reply 1476

StalkeR47

5

Original post by Qari

I forgot to mention keep y the same only talked about x how much do you think i'll get?

The question did not asked for what to keep same, but what to change. you should get 3/3.

Reply 1477

Qari

20

Original post by swagadon

doubling the intensity means doubling the number of photons given out per meter squared,so as there are double the amount of photons( the photons frequency has not changed), there will be double the amount of photoelectrons emitted