Physics, asked by anjalamrssnehadachpa, 1 year ago

NTSE physics question The energy of an electron in n the orbit of a hydrogen atom is given by (a) En = – 13.6 / n2 ev. (b) En = – 13.6 / n3 ev. (c) En = + 13.6 / n2 ev. (d) En = + 13.6 / n3 ev. Please explain! Thanks

Answers

Answered by kvnmurty
2
En = - 13.6 /n²  ev

energy is negative because the electrostatic potential energy is negative and is more than Kinetic energy. 
PE = -  9 *10^9 * Z e * e / r
         r = radius of electron orbit
KE = 1/2 m v²
momentum =  p
lambda = h/p  =>     and  wavelength is an integral multiple of circumference of its orbit.       Lambda  = 2π r * n

velocity v = lambda * nu

centripetal force =     9*10^9 * Z e * e / r²  =  m v² / r

we use the above equations and then  convert a joule into an electron volt we get the answer.

kvnmurty: angular momentum m v r = n h / 2 pi. and n * lambda = 2 pi e
kvnmurty: n lambda = 2 pi r
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