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CBSE Class 12 Physics 2013 Paper

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Question : 12 of 29
Marks: +1, -0
In the ground state of hydrogen atom, its Bohr radius is given as 5.3×10−11m5.3 \times 10^{-11} \text{m}. The atom is excited such that the radius becomes 21.2×10−11m21.2 \times 10^{-11} \text{m}. Find
(i) the value of the principal quantum number and
(ii) the total energy of the atom in this excited state.
Solution:  
(i) r=r0n2r = r_0 n^2
∴ 21.2×10−11  =5.3×10−11×n221.2 \times 10^{-11}\; = 5.3 \times 10^{-11} \times n^2
∴ 4  =n24\; = n^2
n  =2n\; = 2
(ii)   E  =−  13.6eVn2\;E\; = -\;\frac{13.6 \text{eV}}{n^2}
  Or,    E  =−  13.6eV4\;\text{Or,}\;\;E\; = -\;\frac{13.6 \text{eV}}{4}
∴  E  =−3.4eV\therefore \;E\; = -3.4 \text{eV}
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