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  • A gas of identical hydrogen like atoms has some atoms in the lowest energy level A and some atoms in a particular upper energy level B and there are no atoms in any other energy level. The atoms of

A gas of identical hydrogen like atoms has some atoms in the lowest energy level A and some atoms in a particular upper energy level B and there are no atoms in any other energy level. The atoms of the gas make transitions to a higher energy level by absorbing monochromatic light of photon energy 2.7eV. Subsequently, the atoms emit radiation of only six different photon energies. Some of the emitted photons have energy 2.7 eV, some have energy more and some have less than 2.7 eV. The ionization energy for the gas atoms is \mathrm{y} \times 7.2 \mathrm{eV}. Then value of y is:

Option: 1

2


Option: 2

3


Option: 3

4


Option: 4

5


Answers (1)

best_answer

After absorbing 2.7 eV energy. Six different lines are emitted so atoms are excited to n = 4 and it is clear that initially excited state is n = 2
The energy levels of identical hydrogen like atom are given by

E_n=-\frac{B}{n^2} , B being a constant

Given  \mathrm{E}_4-\mathrm{E}_2=2.7 \mathrm{eV}

\therefore-\frac{B}{4^2}-\left(-\frac{B}{2^2}\right)=2.7 \mathrm{eV}

\therefore \text { This gives } B=14.4 \mathrm{eV}

For ionization the transition of electron should be from ground state

(n=1) \text { to } n=\infty

Ionisation energy of identical hydrogen like gas atoms

\Delta E=E_{\infty}-E_1=-\frac{B}{\infty^2}-\left(-\frac{B}{1^2}\right)

\mathrm{B}=14.4 \mathrm{eV}

Posted by

Irshad Anwar

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