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?

 

 
 
 

Answers (1)
S safeer


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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