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
h216π2ma02
c
h232π2ma02
d
h264π2ma02
answer is C.
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Detailed Solution
According to Bohr's modelmvr=nh2π⇒(mv)2=n2h24π2r2⇒KE=12mv2=n2h28π2r2m Also, Bohr's radius for H-atom is, r=n2a0substituting 'r' in Eq. (i) givesKE=h28π2n2a02m when, n=2,KE=h232π2a02m