A hydrogen like atom of atomic number Z is in excited state of quantum number $$3n$$. The maximum energy which can be emitted by atom is $$404$$ eV. If it makes the transition to quantum state n, a photon energy $$40.4$$ eV is emitted. The value of n will be

Question

Infinity Learn · Accepted Answer

in the first case maximum energy is a transition from $$3n$$ to $$1$$  energy released $$13.6$$ x $$Z2112-19n2$$ $$=404$$ $$---1$$   and  in the second case a transition from $$3n$$ to n  $$13.6$$ x $$Z21n2-19n2$$ $$=40.4---2$$ dividing we get $$9n2-18$$ $$=$$ $$10$$   $$n2$$ $$=9$$  $$n=3$$

A hydrogen like atom of atomic number Z is in excited state of quantum number 3n. The maximum energy which can be emitted by atom is 404 eV. If it makes the transition to quantum state n, a photon energy 40.4 eV is emitted. The value of n will be

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