The kinetic energy of an electron in the second Bohr orbit of a hydrogen atom is [a_(0) is Bohr radius]

The kinetic energy of an electron in the second Bohr orbit of a hydrog

1.	The kinetic energy of an electron in the second Bohr orbit of a hydrogen atom is [a_(0) is Bohr radius]
Answer» `(h^(2))/(4pi^(2) m a_(0)^(2))` `(h^(2))/(16pi^(2) m a_(0)^(2))` `(h^(2))/(32pi^(2) m a_(0)^(2))` `(h^(2))/(64 pi^(2) m a_(0)^(2))` SOLUTION :By Bohr's postulate, `MV r= (nh)/(2pi)` `:. v = (nh)/(2pi mr)` `K.E. = (1)/(2) mv^(2) = (1)/(2) m (n^(2) h^(2))/(4pi^(2) m^(2) r^(2))`..(i) But `r = (a_(0) n^(2))/(Z)` for 2ND Bohor orbit of hydrogen, `n = 2, Z = 1` `:. r = (a_(0) xx 4)/(1) = 4a_(0)` Substituting in eqn (i) `K.E. = (1)/(2) (m(2)^(2) h^(2))/(4pi^(2) m^(2) (16a_(0)^(2))) = (h^(2))/(32pi^(2) ma_(0)^(2))`