how is this graph possible? (1 Viewer)

[anonymoushehe]

this from the 2014 HSC physics paper, but I can't understand how this maximum kinetic energy versus frequency curve is possible? wouldnt this graph imply that the work function is 0, which doesnt't make sense to me?

 

[Tony Stark]

this from the 2014 HSC physics paper, but I can't understand how this maximum kinetic energy versus frequency curve is possible? wouldnt this graph imply that the work function is 0, which doesnt't make sense to me?View attachment 48092

By the general form y=mx+c I would that I am just as baffled as you are. Perhaps, the graph depicts energy versus frequency and not maximum kinetic energy? If you refer to the formula sheet, the only difference is there would be no y-intercept as in the graph. But that would only give the energy of the photons.

What a confusing question... https://educationstandards.nsw.edu....-89061beb-0b51-4820-8514-67fe8aac9d78-nbDp1l7 here is the solution provided by BOSTES
Sample answer:
To find intercept
4.1V = 4.1 × 1.602 × 10–19 J of energy required to be supplied by the photon.
= 6.56 × 10–19 J
hf = 6.56 × 10–19 J
6.56 × 10–19
f = 6.626 × 10–34
= 9.9 × 1014
Gradient = same as Al
From graph maximum KE(eV) = 1.2 eV
Frequency = 12.8 Hz
= 12.8 × 1014 Hz

 

[anonymoushehe]

By the general form y=mx+c I would that I am just as baffled as you are. Perhaps, the graph depicts energy versus frequency and not maximum kinetic energy? If you refer to the formula sheet, the only difference is there would be no y-intercept as in the graph. But that would only give the energy of the photons.

What a confusing question... https://educationstandards.nsw.edu....-89061beb-0b51-4820-8514-67fe8aac9d78-nbDp1l7 here is the solution provided by BOSTES
Sample answer:
To find intercept
4.1V = 4.1 × 1.602 × 10–19 J of energy required to be supplied by the photon.
= 6.56 × 10–19 J
hf = 6.56 × 10–19 J
6.56 × 10–19
f = 6.626 × 10–34
= 9.9 × 1014
Gradient = same as Al
From graph maximum KE(eV) = 1.2 eV
Frequency = 12.8 Hz
= 12.8 × 1014 Hz

solutions didnt even make sense to me

 

[Tony Stark]

solutions didnt even make sense to me

The gradient isn't even Planck's constant, unless a degree of accuracy must be considered within the experiment:

 

[coolcat6778]

Old

this from the 2014 HSC physics paper, but I can't understand how this maximum kinetic energy versus frequency curve is possible? wouldnt this graph imply that the work function is 0, which doesnt't make sense to me?View attachment 48092

shitty old syllabus, literally just do relavant vce paper questions instead

Physics - Victorian Curriculum and Assessment Authority

Physics

www.vcaa.vic.edu.au

 

[cheesynooby]

The gradient isn't even Planck's constant, unless a degree of accuracy must be considered within the experiment:

i think the gradient is? wdym

 

[Tony Stark]

i think the gradient is? wdym

As in the gradient should be (E=hf) and (Kmax=hf-phi) h=6.626*10^-34 but rearranging the rise and run of the graph, I get h=6.6775*10^-20

 

[cheesynooby]

As in the gradient should be (E=hf) and (Kmax=hf-phi) h=6.626*10^-34 but rearranging the rise and run of the graph, I get h=6.6775*10^-20

were u forgetting the 10^14 on the x axis

 

[wizzkids]

If you read the whole question from 2014, you will find that the question makes perfect sense.
Go look at the whole question.
2014 Question 26 (b) shows that there is a positive bias applied to the photodetector tube from an external battery.
That means that the positive bias is exactly equal to the work function of the aluminium, measured in eV.