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VCE Stuff => VCE Science => VCE Mathematics/Science/Technology => VCE Subjects + Help => VCE Physics => Topic started by: HERculina on August 28, 2012, 11:34:37 am

Title: Photoelectric effect
Post by: HERculina on August 28, 2012, 11:34:37 am: Has anyone done question 6 exercise 11.4 the dual nature of light.
Can someone explain why/how the max kinetic Energy versus f graph is drawn. I looked at worked solutions but it made me more confused :'(
And what is a trough and a crest in waves?
Thanks.
Title: Re: Photoellectric effect
Post by: HERculina on August 28, 2012, 11:35:52 am: Question is from Heinemann heinemann textbook
Title: Re: Photoelectric effect
Post by: Lasercookie on August 28, 2012, 05:48:28 pm: $E_{kmax} = hf - W = eV_0$
$y = mx + c$
The gradient of the graph is Planck's constant. The magnitude of the y-intercept of the graph is the work function.
The x-axis is frequencies, the y-axis is stopping voltages / energy.

Blue - $f = 7.3 \times 10^{14} Hz$ - $V_0 = 0.9V$

UV - $f = 10.1 \times 10^{14} Hz$ - $V_0 = 2.1V$

So you have two points.

$m = \frac{rise}{run} = \frac{y_2 - y_1}{x_2 - x_1} = \frac{2.1 - 0.9}{(10.1 - 7.3) \times 10^{14}} = 4.28 \times 10^{-15}$

That's fairly close to the value of Planck's constant in eV s. (note that $\frac{eV}{Hz} = \frac{eV}{\frac{1}{s}} = eV s$ )

For the y-intercept, if you're given a graph in the exam, a quick way is to just extrapolate the line and see where it would hit the x-axis. However, we now know the gradient, so we could just sub two points and solve for W

$eV_0 = hf - W$

$2.1 = (4.28 \times 10^{-15})(10.1 \times 10^{14}) - W$

$W = (4.28 \times 10^{-15})(10.1 \times 10^{14}) - 2.1 = 2.28 eV$

edit: forgot about your second question
Quote from: Hercules on August 28, 2012, 11:34:37 am
And what is a trough and a crest in waves?

(http://9-4fordham.wikispaces.com/file/view/wave_crest.gif/243891203/wave_crest.gif)

crest = bit with the maximum, trough = bit with the minimum
Title: Re: Photoelectric effect
Post by: HERculina on August 29, 2012, 04:54:23 pm: Thank you so much laseredd! I finally understand this q. Now :D