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![[Post New]](/templates/default/images/icon_minipost_new.gif) 28 Jul 2007 23:29:13 IST
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please help whether to do these kind of questions by analysing graphs or other way: i am getting confused the freq and intensity of a light source are both doubled. (1)saturation photocurrent remains almost same (2)max k.e. of photoelectrons is doubled a.both true b.1 true 2 false c.1 false 2 true d.both false please reply soon!! aayush
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28 Jul 2007 23:40:24 IST
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I would rather like doing such questions by equations. Graph is also a good idea, but i dont find myself very comfortable with them. As the intensity is doubled, the saturation photoelectric current will increase for sure. Hence i is false For the frequency, Let v represent frequency, KE1 max = hv - hvo If v is doubled KE2 max = 2hv - hvo = 2(hv - hvo) + hvo = 2KE1 + hvo Hence the KE max becomes greater than twice the initial KE with doubling the frequency. Hence ii is also false Hence the answer is d
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HOPE U GOT IT... |
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28 Jul 2007 23:49:24 IST
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but frnd!! ans here given in hcv2 pg no.364 is b...
but thanks for replying!!
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28 Jul 2007 23:50:13 IST
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any more answers please!!!
make it quick
have to go..
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28 Jul 2007 23:58:46 IST
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nooooooooooo saturation current (cut off current or stopping potential all means same) and photoelectric current dont mean the same thing....stopping potential is that min -ive potential given to anode in a photocell fr wich the photoelectric current becomes zero , denoted by Vs , it doesnt depend on intensity of light.......
and abt K.E
let E1 and E2 be the K.E corr. to frequencies @ and 2@ . if Wo be the work func. so h@ = E1 + Wo 2h@ = E2 + Wo dividing 2 = E2 + Wo / E1 + Wo E2 = 2E1 + Wo it means KE will be more than doubled , if freq. is doubled ans will be b
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29 Jul 2007 00:21:57 IST
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ya...i agree with chimanshu...gud xplanation!!
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29 Jul 2007 00:52:29 IST
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Good work Himanshu.
Saturation current and photo-current are two different things. Dont confuse them. While the 2nd statement can easily be proved false by algebric relations. Also these can be proved by graphs.
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Bipin Kumar Dubey
Chemical Dept.
IIT Kharagpur
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29 Jul 2007 18:13:51 IST
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how it can be solved by graphs?????
neways good work...
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