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Question

The magnitude of potential energy of an electron in the second Bohr orbit of a hydrogen atom is:

[a0 is Bohr radius].

A
h24π2ma20
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B
h216π2ma20
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C
h232π2ma20
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D
h264π2ma20
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Solution

The correct option is B h216π2ma20
mvr=2h2Π.....[n=2 for 2nd orbit]
V=hmΠr
KE=12mv2=12×m×h2Π2m2r2=h22Π2mr2
[P.E.]=2KE=h2Π2mr2....[PE=Ke2r & mv2r=Ke2r2] PE=2KE
r= radius of 2nd orbit,
r=ao×n2z=ao×(2)21=4ao
[PE]=h216Π2ma2o

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