If an electron in n=3 orbit of hydrogen atom jumps down to n=2 orbit, the amount of energy released and the wavelength of radiation emitted are
0.85eV,6566˚A
1.89eV,1240˚A
1.89eV,6566˚A
1.5eV,6566˚A
A
1.89eV,1240˚A
B
1.89eV,6566˚A
C
0.85eV,6566˚A
D
1.5eV,6566˚A
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Solution
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E3⟶2=E3−E2E3=−13.69=−1.51eVE2=−13.64=−3.4eV
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