A hydrogen like atom (atomic number Z) is in a higher excited state of quantum number n. The excited atom can make a transition to the first excited state by successively emitting two photons of energy 10.2 eV and 17.0 eV, respectively. Alternatively, the atom from the same excited state can make a transition to the second excited state by successively emitting two photons of energies 4.25 eV and 5.95 eV, respectively Determine the values of n and Z. (lonization energy of H-atom = 13.6 eV)

Explanation

`10.2+17=13.6Z^(2)((1)/(2^(2))-(1)/(n^(2)))`
`27.2=13.6 Z^(2)((1)/(4)-(1)/(n^(2)))`
`4.25+5.95=13.6Z^(2)((1)/(3^(2))-(1)/(n^(2)))`
`10.2=13.6Z^(2)((1)/(9)-(1)/(n^(2)))`
`(27.2)/(10.2)=(27.2)/(10.2)=(1//4-1//n^(2))/(1//9-1//n^(2))`
`(27.2)/(9)-(27.9)/(9)-(10.2)/(4)`
`n^(2)=37impliesn-6`
`27.2=13.6Z^(2)((1)/(4)-(1)/(n^(2)))`
`=13.6Z^(2)((1)/(4)-(1)/(6^(2))=(1)/(4)-(1)/(36)=(9-1)/(36))`
`=13.6Z^(2)xx(2)/(9)`
`Z^(2)=9impliesZ=3`