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Ask iit jee aieee pet cbse icse state board community Community Discussion Question: atomic stucture
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zeeshanp (136)

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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


i can feel the light betray me...............
not a single second left..........
    
zeeshanp (136)

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pls solve and help!!!!!


 


how is it?????


i can feel the light betray me...............
not a single second left..........
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zeeshanp (136)

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isnt anybody getin da answer?


i can feel the light betray me...............
not a single second left..........
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kria (474)

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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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ganesha1991 (1453)

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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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zeeshanp (136)

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rong answer kira


i can feel the light betray me...............
not a single second left..........
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akshay.khare91 (465)

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i am totally agreed wid kria.....

I DONOT FOLLOW THE RULES I MAKE THEM TO FOLLOW ME.
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akshay.khare91 (465)

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the formula lemda = h / underrroot of 2ME is right..

I DONOT FOLLOW THE RULES I MAKE THEM TO FOLLOW ME.
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zeeshanp (136)

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akshay.khare91


 


 


 


is da answer i have mentioned coming?


i can feel the light betray me...............
not a single second left..........
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akshay.khare91 (465)

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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..

I DONOT FOLLOW THE RULES I MAKE THEM TO FOLLOW ME.
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little_genius (295)

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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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kria (474)

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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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vibhu.oct (259)

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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.


VAIBHAV
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vibhu.oct (259)

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The Einfinity= 3.4*1.602*10^-19


sorry, the energy will not be negative.


VAIBHAV
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kria (474)

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