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. The value of n will be

A
2
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B
3
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C
1
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D
4
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Solution

## The correct option is A 2Let the ground state energy (in eV) be E1 Then from the given condition maximum energy of photon that can emit is 240 eV ∴ E2n−E1=204 eV or,E14n2−E1=204 eV or,E1(14n2−1)=204 eV...(i) and E2n−En=40.8 eV or,E14n2−E1n2=E1(−34n2)=40.8 eV −3E14n2=40.8 eV------(ii) From equation (i) and (ii), 1−14n234n2=5⇒154n2=1−14n2⇒4n2=1 or, n=2 Hence, option (B) is correct.

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