Atomic StructureHard

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

Options

A.1
B.2
C.3
D.4

Solution

Let ground state energy (in eV) be E1.
Then from the given condition
E2n -  E1 = 204eV or - E1 = 204eV
⇒ E1 = 204 eV   .....(i)
and E2n - En = 40.8 eV
⇒  = 40.8 eV
From equation (i) and (ii),
= 5 ⇒  n = 2

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