# Use Bohr’s postulates to derive the expressions for the potential and kinetic energy of the electron moving in the nth orbit of the hydrogen atom. How is the total energy of the electron expressed in terms of its kinetic and potential energies?

If the electron moves on a circular path of the radius r with speed V as shown in the figure
$F_{e}= F_{c}$
$\frac{mV^{2}}{r}= \frac{1}{4\pi \varepsilon _{0}}\: \frac{e^{2}}{r^{2}}$
$mV^{2}= \frac{1}{4\pi \varepsilon _{0}}\frac{e^{2}}{r}\rightarrow (1)$
From Bohr's postulate
$mVr= \frac{nh}{2\pi }\rightarrow (2)$
From equations (1) and (2)
$V= \frac{mV^{2}r}{mVr}= \frac{\frac{1}{4\pi \varepsilon _{0}}e^{2}}{\frac{nh}{2\pi }}= \frac{e^{2}}{4\pi \varepsilon _{0}}\times \frac{2\pi }{nh}$
$V= \frac{e^{2}}{2nh\varepsilon _{0}}$
Kinetic Energy

, $KE= \frac{1}{2}mV^{2}$
$= \frac{1}{2}\times m\times \frac{e^{4}}{4n^{2}h^{2}\varepsilon _{0}^{2}}$
$KE= \frac{me^{4}}{8n^{2}h^{2}\varepsilon _{0}^{2}}$
Potential Energy,

$PE= \frac{1}{4\pi \varepsilon _{0}}\cdot \frac{e\times \left ( -e \right )}{r}$
$= \frac{-1}{4\pi \varepsilon _{0}}\frac{e^{2}}{r}$
Referrring to equation (1)
$\frac{-1}{4\pi \varepsilon _{0}}\frac{e^{2}}{r}= -mV^{2}$
$PE= -m\times \left ( \frac{e^{2}}{2nh\varepsilon _{0}} \right )^{2}= \frac{-me^{4}}{4n^{2}h^{2}\varepsilon _{0}^{2}}$
Total energy

$KE+PE$
$= \frac{me^{4}}{8n^{2}h^{2}\varepsilon _{0}^{2}}-\frac{me^{4}}{4n^{2}h^{2}\varepsilon _{0}^{2}}$
$= \frac{me^{4}-2me^{4}}{8n^{2}h^{2}\epsilon _{0}}= \frac{-me^{4}}{8n^{2}h^{2}\varepsilon _{0}}$
Total Energy

$= \frac{-me^{4}}{8n^{2}h^{2}\varepsilon _{0}}$
Total energy = - (kinetic energy)

Potential energy= 2(total energy)

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