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Making use of the uncertainty principle,evaluate the minimum permitted energy of an electron in a hydrogen atom and its corresponding apparent distance from the nucleus. |
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Answer» Solution :HENCE we WRITE `R~Deltar,p~Deltap~ħ//Deltar` Then `E=(ħ^(2))/(2m r^(2))-(e^(2))/(r )` `=(1)/(2m)((ħ)/(r )-(me^(2))/(ħ))^(2)-(me^(4))/(2ħ^(2))` Hence `r_(eff)=(ħ^(2))/(me^(2))=53p m` for the equilibrium state. and then `E=(me^(4))/(2ħ^(2))= -13.6eV` |
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