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Obtain the first Bohr’s radius and the ground state energy of a muonic hydrogen atom [i.e., an atom in which a negatively charged muon (mu^(–)) of mass about 207m_e orbits around a proton]. |
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Answer» Solution :Acording to Bohr MODEL if electron revolvi in orbit on r then `r=(n^(2)h^(2)epsi_(0))/(pi me^(2))` [ `:.` All other terms are constant and energy of electron in orbit of radius r,] `E=-(me^(4))/(8 epsi_(0)^(2)n^(2))` `:. E prop-m ""...(2)` [ `:.` All other terms are constant From equation (1)] `:.(r_(mu))/(r_(e))=(m_(e))/(m_(mu))` where `r_(mu` = orbital radius of muon `r_(e)=` orbital radius of electron `m_(e)=` mass of electron `m_(mu)`= mass of moun `:. r_(mu)=r_(e)xx(m_(e))/(m_(mu))` `0.53xx10^(-10)xx(m_(e))/(207m_(e))` `=2.56xx10^(-13)m` Frm equation (2), `(E_(mu))/(E_(1))=(m_(mu))/(m_(e))` where `E_(mu)=` GROUND state energy of moun `E_(e)=` ground state energy of electron `F_(mu=E_(e)xx(m_(mu))/(m_(e))` `:.E_(mu)=13.6xx207eV` `:.E_(mu)=-2.8152xx10^(3)eV` `:. E_(mu=-2.8KeV` |
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