15. The ratio of the difference in energy between the first and the second Bohr orbit to that between the second and the third Bohr orbit is
Options:
(a) rises continuously
(b) remains unchanged in the process
(c) first rises and then falls to the original position
(d) first falls and then rises to the original position
[ans :d]
Answers
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)
Solution :
For H-atom
E3=−13.632E3=−13.632
=−13.69=−13.69
=−1.5eV=−1.5eV
E2=−13.622E2=−13.622
=−13.64=−13.64
=−3.4eV=−3.4eV
E1=−13.612E1=−13.612
=−13.61=−13.61
=−13.6eV=−13.6eV
Now E2−E1=(−3.4)−(−13.6)E2−E1=(−3.4)−(−13.6)
=13.6−3.4=10.2eV=13.6−3.4=10.2eV
E3−E2=(−1.5)−(3.4)E3−E2=(−1.5)−(3.4)
=3.4−1.5=1.9eV=3.4−1.5=1.9eV
∴E2−E1E3−E2=10.21.9∴E2−E1E3−E2=10.21.9eVeV
=5.4=275
Solution :
For H-atom
E3=−13.632E3=−13.632
=−13.69=−13.69
=−1.5eV=−1.5eV
E2=−13.622E2=−13.622
=−13.64=−13.64
=−3.4eV=−3.4eV
E1=−13.612E1=−13.612
=−13.61=−13.61
=−13.6eV=−13.6eV
Now E2−E1=(−3.4)−(−13.6)E2−E1=(−3.4)−(−13.6)
=13.6−3.4=10.2eV=13.6−3.4=10.2eV
E3−E2=(−1.5)−(3.4)E3−E2=(−1.5)−(3.4)
=3.4−1.5=1.9eV=3.4−1.5=1.9eV
∴E2−E1E3−E2=10.21.9∴E2−E1E3−E2=10.21.9eVeV
=5.4=275
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