Question

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.

Answer 1

$$\begin{gathered} DearStudent, \\ Bohr\text{'}sradius=r_o \\ Radiusafterextication=r_m \\ {r}_m=n{r}_o \\ U\sin ggivendataweget, \\ n(5.3\times {10}^{-11})=(21.2\times {10}^{-11}) \\ \Rightarrow n=\frac{21.2\times {10}^{-11}m}{5.3\times {10}^{-11}m} \\ \Rightarrow n=4 \\ 2)totalenergyintheexcitedstate=\frac{-13.6}{n^2}eV=\frac{-13.6}{4^2}eV=-0.85eV \\ Regards \end{gathered}$$