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CBSE Class 12 Physics 2015 Delhi Set 1 Paper

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Question : 7 of 26
Marks: +1, -0
Show that the radius of the orbit in hydrogen atom varies as n2n^2, where nn is the principal quantum number of the atom.
Solution:  
Showing that the radius of orbit varies as n2n^2
    mv2r=  14πε0  e2r2\;\; \frac{m v^2}{r} = \; \frac{1}{4 \pi \varepsilon_0} \; \frac{e^2}{r^2}
or   mv2r=  14πε0e2\; m v^2 r = \; \frac{1}{4 \pi \varepsilon_0} e^2 .......(i)
mvr=nh2πm v r = \frac{n h}{2 \pi}
m2v2r2=n2h24π2m^2 v^2 r^2 = \frac{n^2 h^2}{4 \pi^2} .......(ii)
Divide (ii) by (i)
  mr  =  n2h24π2×4πε0e2\; m r \; = \; \frac{n^2 h^2}{4 \pi^2} \times \frac{4 \pi \varepsilon_0}{e^2}
  r  =  n2h24π2me24πε0\therefore \; r \; = \; \frac{n^2 h^2}{4 \pi^2 m e^2} \cdot 4 \pi \varepsilon_0
  rn2\therefore \; r \propto n^2
(Give full credit to any other correct alternative method)
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