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Q.

A hydrogen-like atom of atomic number Z is in an excited state of energy quantum number quantum number 2n. It can emit a maximum energy photon of 204 eV. If it makes a transition to ground state , a photon of energy 40.9 eV is emitted. The value of n will be

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a

1

b

2

c

3

d

4

answer is B.

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Detailed Solution

Let ground state energy (in eV) be E1 Then from the given conditionMaximum energy of photon = 13.6 z2 (1 -1∞ ) =13.6 z2 ev Therefore 13.6 z2 =204……………(1) Again 13.6 z2 (1 - 14n2 ) =40.9 …………(2) Solving , n=2
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