The circumference of the second Bohr orbit of an electron in a hydrogen atom is `600 nm` Calculate the potential difference to which the electron has to be Broglie wavelength curresponding to this circumferance

Explanation

`mvr = (nh)/(2pi)`
`2 pi r = (nh)/(mv) = n lambda`
`lambda = (2pi r)/(n) = (600)/(2) = 3000 nm = (h)/(mv)`
`v = (h)/(m lambda)`
If `V_(o)` is voltage
` eV_(0) = (1)/(2) mv^(2) = (1)/(2)m xx (h^(2))/(m^(2) lambda^(2))`
`V_(0) = (h^(2))/(2m lambda^(2) e)`
`= ((6.626 xx 10^(-34) J s)^(2))/(2 xx (9.1 xx 10^(-31 kg)) xx (300 xx 10^(-9)m)^(2))`
`xx 1.6 xx 10^(-19)C`
`= 1.675 xx 10^(-19) V`