consider a neutron and an electron bound to each other due to gravitational force derive an expression for the energy of n-e system

Answers (1)

\\ m_evr = \frac{nh}{2pi}  -(1)

\\ \frac{(Gm_nm_e)}{r^2} = \frac{(m_ev^2)}{r} \\ \Rightarrow \frac{(Gm_n)}{r} =v^2   -(2)

Squaring (2) and dividing it with (1) 

\\ \frac{m_e^2v^2r^2}{v^2} = \frac{n^2h^2r}{4pi^2Gm_n} \\ \Rightarrow m_e^2r= \frac{n^2h^2r}{4pi^2Gm_n} \\ \Rightarrow r= \frac{n^2h^2}{4pi^2Gm_nm_e^2}

\\ \Rightarrow v= \frac{nh}{2pirm_e} \\ \Rightarrow KE= \frac{1}{2m_0v^2} \\ =\frac{1}{2m_0((2piGm_nm_e)^2)/(nh)} \\ \Rightarrow PE = \frac{-Gm_nm_e}{r}

\\ = \frac{-Gm_em_n4pi^2Gm_nm_e^2}{n^2h^2} \\ = \frac{-4pi^2G^2m_n^2m_e^2}{2n^2h^2}

Total energy =KE + PE 
\\= \frac{2a^2G^2m_n^2m_e^2}{n^2h^2}

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