The kinetic energy of an electron in the second Bohr orbit of hydrogen atom: `(a_(0)` is Bohr radius)

Options

A. `(h^(2))/(4pi^(2) ma_(0)^(2))`
B. `(h^(2))/(16pi^(2) ma_(0)^(2))`
C. `(h^(2))/(32pi^(2) ma_(0)^(2))`
D. `(h^(2))/(64pi^(2) ma_(0)^(2))`

Answer

C

Explanation

For Bohr' orbit, angular momentum is
`mvr = (nh)/(2pi), v = (nh)/(2pi m r_(n))` ….(1)
Kinetic energy, `KE = (1)/(2)mv^(2)` …(2)
from (i) and (ii)
`KE =(1)/(2)m xx (n^(2)h^(2))/(4pi^(2) m^(2) r_(n)^(2)) =(n^(2)h^(2))/(8pi^(2) mr_(n)^(2))`
For second Bohr orbit, `n = 2, r_(n) = 4a_(0) r_(n) = a_(0) xx n^(2)` where `a_(0)` is Bohr's radius.
`:. KE = ((2)^(2) (h)^(2))/(8pi^(2) m(4a_(0))^(2)) = (h^(2))/(32 pi^(2) ma_(0)^(2))`