Question

Ionization potential of hydrogen is $13.6$ volt. If it is excited by a photon of energy $12.1eV$, then the number of lines in the emission spectrum will be:

• A. $2$
• B. $3$
• C. $4$
• D. $5$

Answer 1

The correct option is B $3$

In this case, the energy of H-atom after it absorbs photon of energy 12.1eV will be $=-13.6eV+12.1eV=-1.5eV$

Since,

$E_n=-R_H\left(\frac{1}{n^2}\right)$

$n^2=\frac{-R_H}{E_n}=\frac{-2.18\times {10}^{-18}}{-1.5\times 1.6\times {10}^{-19}}=9.08$

$n=3$