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![[Post New]](/templates/default/images/icon_minipost_new.gif) 5 Jul 2008 20:31:02 IST
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if the total enrgy of an electron in hydrofgen like atom in an excited state -3.4ev, then the de broglie wavelength of the electron is..........
ans........ 7.38 x 10^ -10
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i can feel the light betray me...............
not a single second left.......... |
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![[Post New]](/templates/default/images/icon_minipost_new.gif) 5 Jul 2008 20:31:31 IST
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pls solve and help!!!!!
how is it?????
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i can feel the light betray me...............
not a single second left.......... |
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![[Post New]](/templates/default/images/icon_minipost_new.gif) 5 Jul 2008 20:53:56 IST
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isnt anybody getin da answer?
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i can feel the light betray me...............
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![[Post New]](/templates/default/images/icon_minipost_new.gif) 5 Jul 2008 20:55:46 IST
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wll energy =3.4 given
wll use this formula :-

mass of electron put here nd got the answer
correct me if i m wrong
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![[Post New]](/templates/default/images/icon_minipost_new.gif) 5 Jul 2008 20:56:40 IST
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E=h v 3.4 = h v v= 3.4/h
c =lambda * v
c = 3*10^8 h= 6.626 *10^-34 find it from here
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![[Post New]](/templates/default/images/icon_minipost_new.gif) 5 Jul 2008 20:59:01 IST
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rong answer kira
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![[Post New]](/templates/default/images/icon_minipost_new.gif) 5 Jul 2008 21:04:23 IST
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i am totally agreed wid kria.....
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I DONOT FOLLOW THE RULES I MAKE THEM TO FOLLOW ME. |
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![[Post New]](/templates/default/images/icon_minipost_new.gif) 5 Jul 2008 21:04:54 IST
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the formula lemda = h / underrroot of 2ME is right..
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![[Post New]](/templates/default/images/icon_minipost_new.gif) 5 Jul 2008 21:05:49 IST
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akshay.khare91
is da answer i have mentioned coming?
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![[Post New]](/templates/default/images/icon_minipost_new.gif) 5 Jul 2008 21:05:55 IST
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but remember energy in the formula is kinetic energy and kinetic energy = negative of total energy hence u must put 3.4V as energy..
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![[Post New]](/templates/default/images/icon_minipost_new.gif) 5 Jul 2008 21:16:39 IST
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ya the KE = - total energy ... so kria is right... so if the answer still isnt coming then the buk's answer must be wrong...as simple as dat...:)
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![[Post New]](/templates/default/images/icon_minipost_new.gif) 5 Jul 2008 21:20:51 IST
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thanx frnds and u zeeshanp plz confirm the answer once
hey me derived it in this way
e=k.E v=(2EM)^1/2
lmda =h/mv put v here
so ...
get answer
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![[Post New]](/templates/default/images/icon_minipost_new.gif) 5 Jul 2008 21:21:28 IST
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E = -3.4eV = -3.4 * 1.602*10-19 J
This energy is equal to its Kinetic energy.
So, 1/2*mv2=Einfinity
Find v from here.
Put v in lambda=h/mv and find the de Broglie's wavelength.
Correct me, if i am wrong.
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VAIBHAV |
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![[Post New]](/templates/default/images/icon_minipost_new.gif) 5 Jul 2008 21:25:04 IST
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The Einfinity= 3.4*1.602*10^-19
sorry, the energy will not be negative.
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VAIBHAV |
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![[Post New]](/templates/default/images/icon_minipost_new.gif) 5 Jul 2008 21:26:55 IST
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