Question

# A hydrogen-like atom of atomic number Z is in an excited state of quantum number 2n. It can emit a maximum energy photon of 204 eV. If it makes a transition to quantum state n, a photon of energy 40.8 eV is emitted. Find n,Z and the ground state energy (in eV) for this atom. Also, calculate the minimum energy (in eV) that can be emitted by this atom during de-excitation. Ground state energy of hydrogen atom is âˆ’13.6 eV.

A
3,4,10.5 eV.
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B
8,4,10.5 eV.
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C
2,4,10.5 eV.
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D
2,4,15.5 eV.
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Solution

## The correct option is B 2,4,10.5 eV.Given: E2n−E1=204eV, E2n−En=40.8eVSolution:Let us assume that the ground state energy (in eV) be E1 Then from the given condition, we have E2n−E1=204eV orE14n2−E1=204eV or E1(14n2−1)=204eV.....(i) E2n−En=40.8eV or E14n2−E1n2=40.8eV or E1(−34n2)=40.8eV....(ii)From Eqs. (i) and (ii), we get1−14n234n2=5 or 1=14n2+154n2 or 4n2=1 or n =2 From Eq. (ii) E1=−43n2(40.8)eV=−43n2(40.8)eV or E1=−217.6eVE1=−(13.6)Z2 ∴Z2=E113.6=−217.6−13.6=16∴Z=4 Emin=E2n−E2n−1=E14n2−E1(2n−1)2∴Emin=10.58eVHence C is the correct option

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