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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 transition to the first excited state by succesively 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 successively emitting two photons of energies `4.25 eV` and ` 5.95 eV` respectively. Determine the values of n and z

Text Solution

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The correct Answer is:
`n=6,Z=3`
`{:(nrarr2DeltaE=27.2eV(17+10.2)),(nrarr3DeltaE=10.2eV(4.25+5.95.2)):}}E_(3)-E_(2)=(E_(4)-E_(3))^(H)xxZ^(2)`
`17eV= 1.89xxZ^(2)impliesZ=3`
`E_(2)=-3.4xx9=-30.6eV`
`E_(n)-E_(2)=27.2eV`
`E_(n)=27.2+E_(2)=-3.4eV`
`E_(n)=-3.4=-(13.6xx3^(2))/(n^(2))impliesn^(2)=36impliesn=6`
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