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  • In the ground state of hydrogen atom, its Bohr radius is given as . The atom is excited such that the radius becomes <img alt="

In the ground state of hydrogen atom, its Bohr radius is given as 5.3\times 10^{-11}m. The atom is excited such that the radius becomes 21.2\times 10^{-11}m. Find (i) the value of the principal quantum number and (ii) the total energy of the atom in this excited state.

 

 

 

 
 
 
 
 

Answers (1)

(i) r_{n}=r_{0}n^{2}

r_{0} = Bohr radius = 5.39 \times 10^{-11}m

r_{n}=21.2 \times 10^{-11}m

n^{2}=\frac{21.2\times 10^{-11}}{5.3}=4

n = 2

(ii) Total energy

E_{n}=\frac{-13.6}{n^{2}}eV

E_{n}=\frac{-13.6}{4}eV

E_{n}=-3.4eV

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Safeer PP

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