Energy E of a hydrogen atom with principal quantum number n is given by E = −13.6n2eV.The energy of a photon ejected when the electron jumps from n = 3 state to n = 2 state of hydrogen is approximately.
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a
1.5 eV
b
0.85 eV
c
3.4 eV
d
1.9 eV
answer is D.
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Detailed Solution
given En=−13.6n2eVEnergy of photon ejected when electron jumps from n =3 state to n=2 state is given byΔE=E3−E2∴E3=−13.632eV=−13.69eVE2=−13.622eV=−13.64eVSo, ΔE=E3−E2=−13.69−−13.64=1.9eVapproximately