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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 toquantum state n, & photon of energy 40.8 eV is emitted. The value of n is__________

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answer is 2.

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Detailed Solution

Let ground state energy (in eV) be E1Then from the given conditionE2n−E1=204eV or E14n2−E1=204eV⇒ E114n2−1=204eV........(i) and  E2n−En=40.8eV⇒ E14n2−E1n2=E1−34n2=40.8eV......(ii) From Eqs. (i) and (ii), 1−14n234n2=5⇒n=2
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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 toquantum state n, & photon of energy 40.8 eV is emitted. The value of n is__________