InterviewSolution
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InterviewSolution
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If a proton had a radius R and the charge was uniformly distributed, calculate using Bohr theory, the ground state energy ofa H-atom when (i)R=0.1Å, and (ii)R=10Å |
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Answer» Solution :In ground state of hydrogen atom, `mvr_(B)=h, v=h/(mr_(B)), and (mv^(2))/(r_(B))=(-e^(2))/(4pi in_(0)r_(B)^(2))` `:. m/(r_(B))xx(h^(2))/(m^(2)r_(B)^(2))=(e^(2))/(4pi in_(0)r_(B)^(2))` `r_(B)=(4pi in_(0))/(e^(2)) (h^(2))/m=0.51Å` `PE=-(e^(2))/(4pi in_(0)) (1/r_(B))=-27.2eV`. `KE=1/2 mv^(2)=1/2m. (h^(2))/(m^(2)r_(B)^(2))=(h^(2))/(2mr_(B)^(2))=+13.6eV` for special nucleus of radius R, If `R lt r_(B)`, ground state energy remains the same as calculated above. If `R gt gtr_(B)`: the electron moves INSIDE the sphere with radius `r_(B)'` (new Bohr radius) Charge inside `e'=e((r'_(B)^(3))/(R^(3)))` form (i) `r'_(B)=(4pi in_(0))/(e'^(2)) (h^(2))/m=(4pi in_(0)h^(2)(R^(3))^(2))/(me^(2) (r'_(B)^(3))^(2))` `(r'_(B))^(7)=(h^(2))/m((4pi in_(0))/(e^(2)))R^(6)=(0.51Å)R^(6)`.....(i) `r'_(B)=(0.51Å)^(1//7)xxR^(6//7)` If `R=0.1Å`, `r'_(B)=(0.51Å)^(1//7)xx(0.1)^(6//7)=0.126ÅgtR` If `R=10Å`, `r'_(B)=(0.51Å)^(1//7)xx(10)^(6//7)=7.19Å lt R` Now, `K.E. =1/2mv^(2)=(h^(2))/(2mr'_(B)^(2))=(h^(2))/(2mr_(B)^(2))((r_(B)^(2))/(r'_(B)^(2)))=(13.6eV) ((0.51Å)^(2))/((7.19Å)^(2))=0.068eV` `P.E.=+((e^(2))/(4pi in_(0)))((r'_(B)^(2)-3R^(2))/(R^(3)))=(e^(2))/(4pi in_(0)r_(B))[(r_(B)(r'_(B)^(2)-3R^(2)))/(R^(3))]=(27.2eV)[(0.51(7.19^(2)-300))/1000]` `=3.44eV` |
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