Q133 / 200 · NEET 2004 · Physics
Energy $E$ of a hydrogen atom with principal quantum number $n$ is given by $E=\frac{-13.6}{n^2} \mathrm{eV}$. The energy of a photon ejected when the electron jumps for $n=3$ state $n=2$ state of hydrogen is approximately
- A$1.5 \mathrm{eV}$
- B$0.85 \mathrm{eV}$
- C$3.4 \mathrm{eV}$
- Correct answerD$1.9 \mathrm{eV}$
Solution
Energy of photon $=E_2-E_2$
$$
\begin{gathered}
=\frac{13.6}{9}-\left(\frac{-13.6}{9}\right)=\frac{5}{30} \times 13.6 \\
=1.9 \mathrm{eV}
\end{gathered}
$$
Hence, the correct answer is option: 'D'.