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.

CBSE Class 12 Physics 2013 Paper - Question 12

Question : 12 of 29

Marks: +1, -0

In the ground state of hydrogen atom, its Bohr radius is given as

5.3 × 10^{-11} {\m}

. The atom is excited such that the radius becomes

21.2 × 10^{-11} {\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=r_0 n^2

∴

21.2 × 10^{-11}\;=5.3 × 10^{-11} × n^2

∴

4\;=n^2

n\;=2

(ii)

\;E\;=-\;{13.6 {\eV}}/{n^2}

\;\text " Or, "\;\;E\;=-\;{13.6 {\eV}}/{4}

∴ \;E\;=-3.4 {\eV}

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