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Q.

The kinetic energy of an electron in the second Bohr orbit of a hydrogen atom (a0 = Bohr radius)

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a

h24π2ma02

b

h216π2ma02

c

h232π2ma02

d

h264π2ma02

answer is C.

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Detailed Solution

KE=2π2z2me4n2h2for H-atom, Z = 1when n = 2KE=2π2me44h2KE=π2me42h2rn=n2h24π2mze2for H-atom, Z = 1when n = 1r1=h24π2me2but r1=a0a0=h24π2me2π2me2h2=14a0... (2)substituting 2 in 1KE=π2me2h2e22=14a0e22=e28a0multiplying and dividing by a0KE=e2a08a02KE=e28a02h24π2me2substituting a0 value=h232π2ma02

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