anonymous
  • anonymous
Please derive this eqn En = -R (1/n2) where En = energy of the shell R= Rydberg Const n= shell number
IIT study group
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schrodinger
  • schrodinger
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anonymous
  • anonymous
shell/orbit
Mani_Jha
  • Mani_Jha
Ok, let's try this together. The total energy is Kinetic Energy+Potential Energy For Kinetic energy, we've to find the velocity. \[mvr=nh/2\pi\] \[mv ^{2}/r=kq ^{2}/r ^{2}\] Find velocity, then kinetic energy from these equations. Potential energy= \[-kq ^{2}/r\] Try adding the two and see what you get
anonymous
  • anonymous
@mani how did u get those 2 equations? are the pre-defined(q,r) how is centripetal acceleration experienced by the electron towards the nucleus giving it potential energy?

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Mani_Jha
  • Mani_Jha
@Salini, unfortunately people whose username have a space, cannot be pinged here :\ The equation isthe analogue of gravitational motion. In gravitational motion, the centripetal force of revolution of a planet/satellite is provided by the gravitational force of attraction. Here, it is provided by the electrostatic force of attraction due to the nucleus. In any such system where one thing is revolving about another, the energy can't be only kinetic. Do you remember that in Gravitation, potential energy was there to keep the planet bound to the star? Same here. To keep the electron bound to the nucleus, it must possess some negative potential energy. The first equation is a result of Bohr's theory of quantization of angular momentum. Hey @srinidhijha, divide the second equation by the first, and show the result.
anonymous
  • anonymous
i only got to know abt the final energy eqn in ncert text book do u belong to any other board?(this concept is frm 11th right?)
Mani_Jha
  • Mani_Jha
I am in ISC, and even there they have just stated the final result. But you'll find the derivation in some books for competitive examinations. This concept is from the 11th, unless you consider that electricity is taught in class 12. You can watch the MIT video lectures (course 5.112) on atomic structure. They're plainly awesome.
anonymous
  • anonymous
u mean ICSC? if not please specify the full form
Mani_Jha
  • Mani_Jha
ICSE - Indian Certificate of Secondary Education. We had that till class 10. Now, we have ISC - Indian School Certificate. You've never heard of it?
anonymous
  • anonymous
oh icse i wrongly typed as icsc ....oh my ur \board is like 5 times richer than in giving knowledge to students! must be lucky to be in it!
Mani_Jha
  • Mani_Jha
No, actually CBSE is the best. Whether the board teaches us or not, it's good to know a reasoning behind every formula, and every concept you study. Hey srinidhi, please divide the first two equations. Do you want me to do it for you?
anonymous
  • anonymous
no i got the notes thanks
Mani_Jha
  • Mani_Jha
Ok, but just so that the others understand, I am posting the solution: \[mvr=nh/2\pi\] \[v=nh/2pimr\]...........1 \[mv ^{2}/r=kq ^{2}/r ^{2}\] \[v ^{2}=kq ^{2}/mr\].............2 Dividing 2 by 1, \[v=2\pi kq ^{2}/nh\] So energy is: \[mv ^{2}/2=2mpi ^{2}k ^{2}q ^{4}/n ^{2}h ^{2}\] Now, \[2mpi ^{2}k ^{2}q ^{4}/h ^{2}\] is collectively known as the Rydberg's constant(R). So, energy= \[R/n ^{2}\]

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