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  • Consider a hydrogen like atom whose energy in nth excited state is given by <img alt="\mathrm{E_n=-\frac{13.6}{n^2} Z^2}" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cmathrm%7B

Consider a hydrogen like atom whose energy in nth excited state is given by

\mathrm{E_n=-\frac{13.6}{n^2} Z^2}

when this excited atom makes a transition from an excited state to ground state. The most energetic photons have energy \mathrm{E_{\max }=52.224 \mathrm{eV}} and the least energetic photons have energy\mathrm{E_{\min }=1.224 \mathrm{eV}}. Then the atomic number of atom.

Option: 1

2


Option: 2

3


Option: 3

4


Option: 4

5


Answers (1)

best_answer

\mathrm{\mathrm{Max}^{\mathrm{n}} \text { energy is liberated for transition }}

\mathrm{E_n \rightarrow E_1}

\mathrm{\text { and minimum energy for } E_n \rightarrow E_{n-1}}

Hence,

\mathrm{\begin{gathered} \frac{E_1}{n^2}-\frac{E_1}{12}=52.224 \mathrm{eV} \\ \text { and } \frac{E_1}{n^2}-\frac{E_1}{(n-1)^2}=1.224 \mathrm{eV} \end{gathered}}

Solving we get,

\mathrm{\mathrm{E}_1=-54.4 \mathrm{eV}}

and n = 5
hence,

\mathrm{\begin{aligned} &E_1=-\frac{13.6 z^2}{12}=-54.4\\ &Z=2 \end{aligned}}

 

Posted by

Shailly goel

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