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A hydrogen like atom (atomic number Z) i...

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` and `17.00 eV` respectively. Alternatively, the atom from the same axcited state can make a transition to the second excited state by successively emitting two photons of energy `4.25 eV` and `5.95 eV` respectively. Determine the values of n and Z `("ionisation energy of hydrogen atom"=13.6 eV)`. Given answer `=n+Z`.

Text Solution

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The correct Answer is:
9
`27.2xx1.602xx10^(-12)=R_(H)z^(2)hc[1/(2^(2))-1/(n^(2))]`……..(1)
`10.2xx1.602xx10^(-12)=R_(H)z^(2)hc[1/(3^(2))-1/(n^(2))]`…….(2)
After solving (1) and (2)
`n=6, z=3`
`n+z=9`
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