Solution Rdius of nth Bohr orbit rn =ε0 n2h2π m Ze2For the first orbit of Hydrogenn = 1, Z= 1r1 =ε0 h2π m e2r1 =8 85×10-12×6 62×10-34×6 62×10-343 14×9 1×10-31×1 6×10-19×1 6×10-19r1 = 0 528×10-10 mr1 = 0 528 Angstromrn =n2Z×0 528r2 = 4× 0 528 Angstrom = 2 116 Angstrom

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Solution

R d i u s o f n^{t h} B o h r o r b i t r_{n} = \frac{ε_{0} n^{2} h^{2}}{π m {Ze}^{2}} F o r t h e f i r s t o r b i t o f H y d r o g e n n = 1, Z = 1 r_{1} = \frac{ε_{0} h^{2}}{π m e^{2}} r_{1} = \frac{8.85 \times 10^{- 12} \times 6.62 \times 10^{- 34} \times 6.62 \times 10^{- 34}}{3.14 \times 9.1 \times 10^{- 31} \times 1.6 \times 10^{- 19} \times 1.6 \times 10^{- 19}} r_{1} = 0.528 \times 10^{- 10} m r_{1} = 0.528 A n g s t r o m r_{n} = \frac{n^{2}}{Z} \times 0.528 r_{2} = 4 \times 0.528 A n g s t r o m = 2.116 A n g s t r o m

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