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equation total energy for electrons:$$Et=Ek + Ep$$$$=1/2mv^2 + eV$$ Now equation for velocity of electrons at any orbital is:$$v=e/√4πεmr$$ and electric potential $$V=e^2/4πεr^2$$ Plugging in the values and after simplfying we get,$$E=-me^4/8n^2h^2ε^2$$From here we can write $$E∝1/n^2$$So the higher the number of orbital, the less the energy should be. Then why higher orbitals require higher energy in electrons?

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    $\begingroup$ Just to be clear, you've arbitrarily removed a minus sign, after which you're puzzled that things go the wrong way around? $\endgroup$ – Emilio Pisanty Sep 18 '17 at 12:53
  • $\begingroup$ actually no. If I multiply the rydberg's equation with planck's constant then I get change in energy when electron is moving from one orbit to another. That equation clearly shows that electron would require energy to move to a higher orbital. But the equation I mentioned shows higher orbit means lower energy. I am just confused with the contradiction that I am getting. $\endgroup$ – Sami Sep 18 '17 at 12:58
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    $\begingroup$ No, it doesn't. The correct version is $E\propto -1/n^2$, which increases with increasing $n$, i.e. the higher the number of the orbital, the higher the energy. $\endgroup$ – Emilio Pisanty Sep 18 '17 at 13:01
  • $\begingroup$ But if I want to compare the energy of electrons of two different shells, and so find E1/E2 , that would be equal to (n2/n1)^2. The negative signs will be gone. How is that wrong? $\endgroup$ – Sami Sep 18 '17 at 13:10
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we can write $$E∝1/n^2$$So the higher the number of orbital, the less the energy should be.

should read

we can write $$E∝ - 1/n^2$$.
So the higher the number of orbital, the less negative the energy should be

ie the energy is more positive or higher.

Going from $2$ to $3$

$E_{2\rightarrow 3} \propto \left ( -\dfrac {1}{3^2}\right )- \left ( -\dfrac {1}{2^2}\right ) = + \dfrac{5}{36} $

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