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)ma_(0)^(2))`
`(h^(2))/(16pi^(2) ma_(0)^(2))`
`(h^(2))/(32pi^(2) ma_(0)^(2))`
`(h^(2))/(64pi^(2) ma_(0)^(2))`

Solution :`mvr= (nh)/(2PI), mv= (nh)/(2pi r) and KE = (1)/(2)mv^(2)= (m^(2)v^(2))/(2M) or E= (n^(2)h^(2))/(4pi^(2) r^(2))xx (1)/(2m)`
use `r= n^(2) a_(0) (a_(0)=r_(1))`


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