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

Question

Infinity Learn · Accepted Answer

Let ground state energy (in$$ $$eV) be $$E1Then$$ from the given $$conditionE2n$$−$$E1=204eV$$  ⇒ $$E14n2$$−$$E1=204eV$$ ⇒$$E114n2$$−$$1=204eV$$.......(i)  and   $$E2n$$−$$En=40.8eV$$ ⇒$$E14n2$$−$$E1n2=E1$$−$$34n2=40.8eV$$........(ii)  From equations (i) and (ii),  $$1$$−$$14n234n2=5$$ ⇒$$n=2$$

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

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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, a photon of energy 40.8 eV is emitted. The value of n will be

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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 to
quantum state n, a photon of energy 40.8 eV is emitted. The value of n will be