Q.
The frequency of radiation emitted when the electron falls from n=4 to n=1 in a hydrogen atom will be (Given ionisation energy of H=2.18×1018Jatom−1 and
h=6.625×10−34Js)
12440
227
AIPMTAIPMT 2004Structure of Atom
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Solution:
Ionisation energy of H=2.18×10−18Jatom−1 ∴E1 (Energy of Ist orbit of H-atom) =−2.18×10−18Jatom−1 ∴En=n2−2.18×10−18Jatom−1 Z=1 for H-atom ΔE=E4−E1 =42−2.18×10−18−12−2.18×10−18 =−2.18×10−18×[421−121] ΔE=−2.18×10−18×−1615 =+2.0437×10−18Jatom−1 ∴v=hΔE =6.625×10−34Js2.0437×10−18Jatom −1 =3.084×1015s−1 atom −1