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  • Consider a hydrogen like atom whose energy in excited state is given by <img alt="\mathrm{E}_{\mathrm{n}}=-\

Consider a hydrogen like atom whose energy in n^{\text {th }} excited state is given by
\mathrm{E}_{\mathrm{n}}=-\frac{13.6}{\mathrm{n}^2} \mathrm{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}. The atomic number of atom is

Option: 1

2


Option: 2

3


Option: 3

4


Option: 4

5


Answers (1)

best_answer

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

                   \mathrm{E}_{\mathrm{n}} \rightarrow \mathrm{E}_1

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

Hence,

\frac{E_1}{n^2}-\frac{E_1}{12}=52.224 \, \, e V

\text { and } \frac{E_1}{n^2}-\frac{E_1}{(n-1)^2}=1.224 \, \, \mathrm{eV}

Solving we get,

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

\text { and } n=5

hence,

E_1=-\frac{13.6 Z^2}{12}=-54.4

\mathrm{Z}=2

Posted by

vinayak

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