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Author Topic: Photoelectric effect  (Read 749 times)  Share 

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HERculina

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Photoelectric effect
« on: August 28, 2012, 11:34:37 am »
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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.
« Last Edit: August 28, 2012, 11:36:35 am by Hercules »
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HERculina

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Re: Photoellectric effect
« Reply #1 on: August 28, 2012, 11:35:52 am »
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Question is from Heinemann heinemann textbook
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Lasercookie

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Re: Photoelectric effect
« Reply #2 on: August 28, 2012, 05:48:28 pm »
+1


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 - -

UV  - -

So you have two points.



That's fairly close to the value of Planck's constant in eV s.  (note that )

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







edit: forgot about your second question
And what is a trough and a crest in waves?



crest = bit with the maximum, trough = bit with the minimum
« Last Edit: August 28, 2012, 05:58:05 pm by laseredd »

HERculina

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Re: Photoelectric effect
« Reply #3 on: August 29, 2012, 04:54:23 pm »
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Thank you so much laseredd! I finally understand this q. Now :D
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