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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a
2
b
3
c
4
d
5
answer is B.
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Detailed Solution
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