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NCERT Class XII Chapter
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Question : 8 of 18
Marks: +1, -0
The radius of the innermost electron orbit of a hydrogen atom is 5.3 × 10−1110^{-11} m. What are the radii of the n = 2 and n = 3 orbits?
Solution:  
Radius of innermost electron, r = n2h2ε0πme2\frac{n^2 h^2 \varepsilon_0}{\pi m e^2}
For n = 1 , r1r_1 = h2ε0πme2\frac{h^2 \varepsilon_0}{\pi m e^2} = 5.3 × 10−1110^{-11} m
For n = 2, r2r_2 = (2)2r1(2)^2 r_1 = 2.12 × 10−1010^{-10} m
For n = 3, r3r_3 = (3)2r1(3)^2 r_1 = 4.77 × 10−1010^{-10} m.
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