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Q.

The kinetic energy of an electron in the second Bohr orbit of a hydrogen atom is [a0 is Bohr radius]:

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a

h24π2ma02

b

h232π2ma02

c

h264π2ma02

d

h216π2ma02

answer is C.

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Detailed Solution

We know that;

mvr=nh2π P=nh2πr;KE=P22m=n2h28π2mr2Also, r=a0×n2KE=h28π2mn2a02; Put n=2

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