Atomic StructureEasy
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
Then from the given condition
E2n - E1 = 204eV or
- E1 = 204eV⇒ E1
= 204 eV .....(i)and E2n - En = 40.8 eV
⇒
= 40.8 eVFrom equation (i) and (ii),
= 5 ⇒ n = 2Create a free account to view solution
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