The acceleration of electron in Bohr's `1^(st)` orbit is given by

Options

A. `h/(4pi m^(2) r^(3))`
B. `h/(4pi m^(2) r)`
C. `h^(2)/(4pi^2m^(2) r^(3))`
D. `h/(4 pi m r)`

Answer

C

Explanation

`a_(n)=(pi me^(5))/(4 epsi_(0)^(2)n^(4)h^(4))`
`a_(1)=(pi me^(6))/(4 epsi_(0)^(3) h^(4))`
`=(pi me^(6))/(4 epsi_(0)^(2)h^(4))xx(pi^(2)m^(2)h^(2))/(pi^(2) m^(2) h^(2))`
`=(pi^(3)m^(3)e^(6))/(epsi_(0)^(3) h^(6))xxh^(2)/(4pi^(2) m^(2))`
`=h^(2)/(4pi^(2) m^(2) r^(3))" "( :' r_(1)^(3)=(epsi_(0)^(3)h^(6))/(pi^(3)m^(3)e^(6)))`