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A hydrogen like atom with atomic number 'Z' is in higher excited state of quantum number 'n'. This excited state can make a transition to the first excited state by successively emitting two photons energies 10 eV and 17 eV respectively. Alternatively, the atom from the same excited state can make a transition to the second excited state by emitting two photons of energies 4.25 eV adn 5.95 eV respectively. Calculate the values of 'n' and 'Z'

Answer» <html><body><p></p>Solution :For the given <a href="https://interviewquestions.tuteehub.com/tag/excited-2625787" style="font-weight:bold;" target="_blank" title="Click to know more about EXCITED">EXCITED</a> state, `n_(2) = n` <br/> 1st excited state means, `n_(1) =2` <br/> 2nd excited state means, `n_(1) =3` <br/> <a href="https://interviewquestions.tuteehub.com/tag/energy-15288" style="font-weight:bold;" target="_blank" title="Click to know more about ENERGY">ENERGY</a> difference between n = 2 and n = 3 will be `= (10 + <a href="https://interviewquestions.tuteehub.com/tag/17-278001" style="font-weight:bold;" target="_blank" title="Click to know more about 17">17</a>) - (4.25 + 5.95) = 16.8 eV` <br/> `:. 16.8 = 13.6 ((1)/(2^(2)) - (1)/(3^(2))) Z^(2) = 13.6 xx (5)/(36) xx Z^(2)` <br/> or `Z^(2) = (16.8)/(13.6) xx (36)/(5) = 9 or Z = 3` <br/> Energy difference between `n_(2) = n and n_(1) = 2` will be <br/> `= 10 + 17 eV = 27 eV` <br/> `:. 27 = 13.6 ((1)/(2^(2)) - (1)/(n^(2))) xx 3^(2)` <br/> or `3 = 13.6 ((1)/(4) - (1)/(n^(2))) or (1)/(4) - (1)/(n^(2)) = (3)/(13.6) or (1)/(n^(2)) = (1)/(4) - (3)/(13.6) = (13.6 - <a href="https://interviewquestions.tuteehub.com/tag/12-269062" style="font-weight:bold;" target="_blank" title="Click to know more about 12">12</a>)/(4 xx 13.6)` <br/> or `n^(2) = (4 xx 13.6)/(1.6) or n = 6` <br/> <img src="https://d10lpgp6xz60nq.cloudfront.net/physics_images/PR_CHE_01_XI_C02_E03_001_S01.png" width="80%"/></body></html>


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