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A hydrogen-like atom (atomic number Z) is in a higher excited state of quantum number n. This excited atom can make a transition to the first excited state by successively emitting two photons of energies 10.20 eV and 17.00 eV respectively.Alternatively the atom from the same excited state can make a transition to the second excited state by snccessively emitting two photons of energies 4.25 eV and 5.95 eV respectively. Determine the values of n and Z (ionization energy of hydrogen atom = 13.6 eV)

Answer» <html><body><p>n=5<br/>z=2<br/>n=6<br/>z=3</p>Solution :<img src="https://doubtnut-static.s.llnwi.net/static/physics_images/AKS_ELT_AO_CHE_XI_V01_A_C01_E03_124_S01.png" width="80%"/> <br/>`E= (-2^2)/(n^2) (13.6eV)` <br/> `impliesR.hc.z^2 (1/<a href="https://interviewquestions.tuteehub.com/tag/9-340408" style="font-weight:bold;" target="_blank" title="Click to know more about 9">9</a> - 1/n^2) = 27.2 ….. (1)` <br/>`impliesR.hc.z^2 (1/9 - 1/n^2) = 10.2 eV…(2)` <br/>`((1))/((2))= (((n^2 - <a href="https://interviewquestions.tuteehub.com/tag/4-311707" style="font-weight:bold;" target="_blank" title="Click to know more about 4">4</a>)/(4n^2)))/(((n^2 - 9 )/(9n^2))) = (27.2)/(10.2) = 2.66` <br/>`9(n^2 - 4)xx 4 (n^2 - 9) (2.66)` <br/>`implies9n^2 - <a href="https://interviewquestions.tuteehub.com/tag/36-309156" style="font-weight:bold;" target="_blank" title="Click to know more about 36">36</a> = 10 . 66 n^2 - 96` <br/>`therefore 1.66 n^2 = 60 impliesn^2 =(60)/(1.66) = 36`</body></html>


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