Question

The frequency of the spectral line obtained when the electron in $n=3$ of Hydrogen atom drops to the fround state is:

• A. $2.925\times {10}^{15}$Hertz
• B. $2.925\times {10}^{13}$Hertz
• C. $2.925\times {10}^{14}$Hertz
• D. $36559\times {10}^{10}$Hertz

Answer 1

The correct option is A $2.925\times {10}^{15}$Hertz

Energy difference= $13.6[\frac{1}{(1{)}^2}-\frac{1}{(3{)}^2}]eV$

$=12.08eV$

$=12.08\times 1.6\times {10}^{-19}J$

$=1.93\times {10}^{-18}J$

$hv=1.93\times {10}^{-18}J$

$\therefore v=\frac{1.93\times {10}^{-18}}{6.63\times {10}^{-34}}=2.925\times {10}^{15}$ Hertz