A hydrogen atom in its first excited state absorbs a photon of energy x $\times$ 10$-$2 eV and excited to a higher energy state where the potential energy of electron is $-$1.08 eV. The value of x is ______________.
Answer (integer)
286
Solution
As, $E_{n}=\frac{P \cdot E_{n}}{2}=-\frac{1.08}{2}=-0.544$
<br/><br/>
$$
\begin{aligned}
& \text { So, } \Delta \mathrm{E}, \mathrm{E}_{\mathrm{f}}-\mathrm{E}_{\mathrm{i}}=-0.544-\left(-\frac{13.6}{2^{2}}\right)=3.4-0.544 \\\\
& \approx 2.86 \mathrm{eV}=286 \times 10^{-2} \mathrm{eV}
\end{aligned}
$$
About this question
Subject: Physics · Chapter: Atoms and Nuclei · Topic: Bohr's Model of Hydrogen Atom
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