Question

An $H$ -atom in the ground state is excited by monochromatic radiation of photon energy $\text{13.056} \, eV$ . The number of emission lines will be (given its ionisation energy is $\text{13.6} \, eV$ )

• A. one
• B. two
• C. four
• D. ten

Answer 1

Answer: (D)

If $n$ , is the value of excited state, then
number of spectral lines $= \frac{\text{n} \left(\text{n} - 1\right)}{2}$
Energy of electron in ground state= – (ionisation energy) $=-\text{13.6}eV$
Final energy $\text{-13.6+13.056=-0.544 eV}$
If $n$ be the value of excited state, then
$\text{-}\frac{\text{13.6}}{n^{2}}=-\text{0.544}$
$\Rightarrow \text{n}^{2}=25$
$\Rightarrow \text{n}=5$
Now, the number of spectral lines are
$=\frac{\text{n} \left(\text{n} - 1\right)}{2}=\frac{5 \times 4}{2}=10\text{ lines}$