A hydrogen like atom has one electron revolving round a stationary nucleus. If the energy required to excite…
A hydrogen like atom has one electron revolving round a stationary nucleus. If the energy required to excite the electron from the 2 nd orbital to 3 rd orbit is \(47.2 \mathrm{eV}\), find the atomic number of the given atom.
3
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5
6
Solution
The energy of \(n\)th orbit is given as
\(E_n=\frac{-R h C z^2}{n^2}\)
For hydrogen atom, \(z=1\) and \(n=1\), hence
\(E_1=\frac{-R h C \cdot 1^2}{1^2}=-R h C\)
We know that, \(R h C=13.6 \mathrm{eV}\)
\(\therefore \quad E_1=-13.6 \mathrm{eV}\)
For \(n=2\),
\(\begin{aligned}
& E_2=\frac{-R h C z^2}{2^2}=\frac{-13.6 z^2}{4} \\
\Rightarrow \quad & E_2=\frac{-13.6 z^2}{4} \quad \ldots (i)
\end{aligned}\)
Similarly, for \(n=3\),
\(E_3=\frac{-13.6}{9} z^2 \quad \ldots (ii)\)
Given,
\(\begin{array}{ll}
\Rightarrow & \frac{-13.6}{9} z^2-\left(\frac{-13.6 z^2}{4}\right)=47.2 \\
\Rightarrow & 13.6 \times \frac{5}{36} z^2=47.2 \\
\Rightarrow & z^2=24.98 \\
\Rightarrow & z^2 \simeq 25 \\
\Rightarrow & z=5
\end{array}\)