Edexcel Unit 2: Physics at Work ~9th June 2014

OP

help?

The photoelectric effect shows that the release of an electron from the surface of a metal is only dependent on the frequency, and the number of electrons released per second is only dependent on the intensity. This can only be explained by the particle nature of light since photons are packets of energy (quanta) where each photon gives one electron all its energy, which can be used by the electron to escape. Also, because e=hf, the energy of the photon hitting the electron is determined only by the frequency.

If the light was behaving as a wave, then the intensity would determine if the electron was released or not since a wave would be a continuous stream of energy that would "build up" in the electron before it was released, but this doesn't happen as the electrons are released instantaneously (with no time delay). -I think this paragraph's correct but if it isn't please correct me.

Reply 21

BP_Tranquility

OP

5

Hmm..
http://www.edexcel.com/migrationdocuments/QP%20GCE%20Curriculum%202000/June%202012%20-%20QP/6PH02_01_que_20120525.pdf

For question 11a, in the table, the switch combination A open B closed has a total resistance of R---why is it R and not 2R?

Reply 22

username1411983

1

Original post by BP_Tranquility

Hmm..
http://www.edexcel.com/migrationdocuments/QP%20GCE%20Curriculum%202000/June%202012%20-%20QP/6PH02_01_que_20120525.pdf

For question 11a, in the table, the switch combination A open B closed has a total resistance of R---why is it R and not 2R?

Not 2R because the current flows through the path which has least resistance. When current reaches the junction between the two resistors it tends to flow through switch B rather than resistor R :biggrin:

Reply 23

BP_Tranquility

OP

5

Original post by Roflwaffles

Not 2R because the current flows through the path which has least resistance. When current reaches the junction between the two resistors it tends to flow through switch B rather than resistor R :biggrin:

Oh, I didn't know that :redface:

-is this circuit a potential divider by any chance? I thought that current flowed throughout the entire circuit (e.g. if you put two bulbs in parallel with a cell, then both light up but with different brightnesses)

Also, do you know of any good websites/resources to revise from? The specification seems a bit basic and not in depth :rolleyes:

(edited 9 years ago)

Reply 24

arkajyotig

2

Original post by sabphysics

new question!!

the answer is A = 1/4 R.
take the formula:- resistance=(resistivity X lenght)/crossectional area.
Before replacing:
R₁= (resistivity X Lenght)/(pi X (d²)/4)
R₁= 4(resistivity X Lenght)/(pi X (d²))

After replacing:
R₂= (resistivity X Lenght)/(pi X (d²))

4R₂=R₁therefore R₂=(1/4)R₁

Reply 25

arkajyotig

2

Original post by sabphysics

help?

observation 1:- EM radiation should be of minimum required frequency(threshold frequency) for the emmision of a photoelectron. light strikes the metal surface in the form of small packets of energy known as photons which has the energy E=hf. Threshold frequency is when hf=work function(W).thus minimum value for frequency(f) is required for hf to be equal to W.
observation 2:- no. of photoelectrons emitted depends upon the intensity of light used and not the frequency. this is because increase in intensity of light will cause more no. of photons striking the metal surface per second. one photon causes emission of one photoelectron. thus more photoelectrons will be released per second.
observation 3:- max. K.E of photoelectron depends on frequency and not intensity. according to Einstein's equation:- hf-W=K.E
since h(planck's constant) and W(work function) are both constant, increasing frequency will lead to increase in K.E of photoelectron.
observation 4:- no time delay between photon striking metal surface and emmision of photoelectron. if light was a wave, there would be a time delay as energy has to build up before emission of photoelectron which is not observed thus light is not a wave.

Reply 26

username1411983

1

Original post by BP_Tranquility

Oh, I didn't know that :redface:

-is this circuit a potential divider by any chance? I thought that current flowed throughout the entire circuit (e.g. if you put two bulbs in parallel with a cell, then both light up but with different brightnesses)

Also, do you know of any good websites/resources to revise from? The specification seems a bit basic and not in depth :rolleyes:

Nah! sorry. I am doing CIE so no idea about edexcel websites :smile:

Reply 27

Necrosyrtes

4

Original post by arkajyotig

the answer is A = 1/4 R.
take the formula:- resistance=(resistivity X lenght)/crossectional area.
Before replacing:
R₁= (resistivity X Lenght)/(pi X (d²)/4)
R₁= 4(resistivity X Lenght)/(pi X (d²))

After replacing:
R₂= (resistivity X Lenght)/(pi X (d²))

4R₂=R₁therefore R₂=(1/4)R₁

How to do it (and questions like it) if you are lazy:

R = \frac{\rho l}{A}

R \propto \frac{1}{A}

since

\rho

and

l

are held constant in this case
Also

A \propto D^{2}

therefore

R \propto \frac{1}{D^{2}}

Since

D

doubles, substitute 2 into

\frac{1}{D^{2}}

, finding that the new resistance is

\frac{1}{4}R

This looks like a long way to do it, but once you get the hang of it, I find it takes a bit less time, and is easier to think about. It lets me have that extra 20s I might need to answer another question, but if I have time left at the end, I'd definitely check this one - I find it's quite easy to make mistakes if you aren't careful.

Also, you use d in your formulas, which implies that

A = \pi D^{2}

, which isn't true. It doesn't change the answer at all, but could be quite confusing for anyone reading. If you substituted d for r, your process would make more sense. As I said, it makes no difference to the answer - if you half d, you half r, so the proportions stay the same, but to avoid confusion I think r should be used.

Reply 28

arkajyotig

2

Original post by Necrosyrtes

How to do it (and questions like it) if you are lazy:

R = \frac{\rho l}{A}

R \propto \frac{1}{A}

since

\rho

and

l

are held constant in this case
Also

A \propto D^{2}

therefore

R \propto \frac{1}{D^{2}}

Since

D

doubles, substitute 2 into

\frac{1}{D^{2}}

, finding that the new resistance is

\frac{1}{4}R

This looks like a long way to do it, but once you get the hang of it, I find it takes a bit less time, and is easier to think about. It lets me have that extra 20s I might need to answer another question, but if I have time left at the end, I'd definitely check this one - I find it's quite easy to make mistakes if you aren't careful.

Also, you use d in your formulas, which implies that

A = \pi D^{2}

, which isn't true. It doesn't change the answer at all, but could be quite confusing for anyone reading. If you substituted d for r, your process would make more sense. As I said, it makes no difference to the answer - if you half d, you half r, so the proportions stay the same, but to avoid confusion I think r should be used.

ohh anyways thnx for making it clear in case some people dont understand. i didnt know how to type out formulas lol:P

(edited 9 years ago)

Reply 29

Kaksha-97

Hey guys,

I've been revising using old salters horners papers from 2007 onwards and they've been quite helpful and they seem to be much friendlier than the normal edexcel papers in my opinion.

I came across a question about spectral lines, I was wondering where energy comes from when electrons are moved from their ground state to a higher energy level and emit photons? I'm still confused as to whether this is the photoelectric effect or not.

It's question 3 in PSA1 Jan 2007 6751/01

Reply 30

Freddy-Francis

9

Original post by Kaksha-97

Hey guys,

I've been revising using old salters horners papers from 2007 onwards and they've been quite helpful and they seem to be much friendlier than the normal edexcel papers in my opinion.

I came across a question about spectral lines, I was wondering where energy comes from when electrons are moved from their ground state to a higher energy level and emit photons? I'm still confused as to whether this is the photoelectric effect or not.

It's question 3 in PSA1 Jan 2007 6751/01

The most common way for electrons to move to an excited state is by absorption of electromagnetic radiation. When an electron absorbs this radiation, it takes in the energy that the wave contains and this causes it to move to a higher energy state. They come back to ground state at the first possible opportunity because in the excited state they have lots of potential energy. Electrons can make different jumps in energy levels when it loses this potential energy (ie. it can go directly to the ground state, or it can go to lower energy levels and then the ground state). When they move to lower energy levels, the electrons emit the electromagnetic radiation that was absorbed - the frequency of the waves emitted depends on the electron's jumps between energy levels as it returns to ground state.

Reply 31

Kaksha-97

Original post by Freddy-Francis

The most common way for electrons to move to an excited state is by absorption of electromagnetic radiation. When an electron absorbs this radiation, it takes in the energy that the wave contains and this causes it to move to a higher energy state. They come back to ground state at the first possible opportunity because in the excited state they have lots of potential energy. Electrons can make different jumps in energy levels when it loses this potential energy (ie. it can go directly to the ground state, or it can go to lower energy levels and then the ground state). When they move to lower energy levels, the electrons emit the electromagnetic radiation that was absorbed - the frequency of the waves emitted depends on the electron's jumps between energy levels as it returns to ground state.

Thank you, that made things much clearer :smile:

Reply 32

Freddy-Francis

9

Original post by Kaksha-97

Thank you, that made things much clearer :smile:

Happy to help :colondollar:

Reply 33

jtbteddy

5

What exactly is a spectra?

Posted from TSR Mobile

Reply 34

fictioned

1

I feel so scared. Ugh. I cant wait to join this board after tomorrow

Posted from TSR Mobile

Reply 35

Harmanosa

1

Well, I guess it's only this exam left now. I'm gonna be going through the specification points for unit 2 like I did with unit 1 and writing a set of notes for every single point. When I went through the examiner's report on unit 1 physics june 2013 it seemed like they thought we didn't know some basic stuff from the specification, and I'm guessing it'd be an idea just to check it all anyway.

Reply 36

13142vp

Original post by jtbteddy

What exactly is a spectra?

Posted from TSR Mobile

spectra is the arrangement according to wavelengths. so for visible light the spectrum goes from 4-7 x 10^-7m. For observation a spectroscope is used. Emission would produce colored lines and absorption would produce dark lines. a continuous is produced when all wavelengths are present. Emission occurs when electrons in atomic orbit move from high to low energy levels. Absorption occurs when electrons move ground state/low energy levels to high energy levels/excited state.
Hope that is clear :smile:

Reply 37

jtbteddy

5

Original post by 13142vp

spectra is the arrangement according to wavelengths. so for visible light the spectrum goes from 4-7 x 10^-7m. For observation a spectroscope is used. Emission would produce colored lines and absorption would produce dark lines. a continuous is produced when all wavelengths are present. Emission occurs when electrons in atomic orbit move from high to low energy levels. Absorption occurs when electrons move ground state/low energy levels to high energy levels/excited state.
Hope that is clear :smile:

Thank youu :smile:

) And how would i define spectra for a one mark question?