Question

Derive the expression for total energy of electron in $n^{th}$ Bohr orbit and show that -
$E_{n}\propto\frac{1}{n^{2}}$

Answer 1

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Answer 2

Answer:

$T..E \alpha \dfrac{1}{n^{2} }$

Explanation:

Since we know that when electron is revolving in the orbit it experience a centripetal forec as well as force of attraction between revolving electron and nucleous and both forces are equal.

$F_{centripetal} =F_{attraction}$

$\dfrac{mv^{2} }{r} = \dfrac{Kze^{2} }{r^{2} }$......................(1)

From (1) we get K..E

$K.E. = \dfrac{Kze^{2} }{2r}$

We know the potenial energy of a revolving electron is given as

$P.E = \dfrac{-Kze^{2}}{r}$

So, T.E = K.E.+ P.E.

$T.E. = \dfrac{-Kze^{2} }{2r}$.....................(2)

We know the expression for r

$r= \dfrac{n^{2}h^{2} }{4\pi^{2}kmze^{2}}$.........................(3)

From (2) and (3) we get,

$T..E \alpha \dfrac{1}{n^{2} }$