According to Bohr's theory,the electronic energy of a atom in the nth orbit is given by `E_(n) = (-2.17 xx 10^(-18))/(n^(2)) J`
Calculate the longest wavelength of light that will be needed in remove an electron the third Bohr orbit of `He^(+)

Answer

A, B, D

Explanation

`E_(n) = (-21.76 xx 10^(-19))/(n^(2)) J`
For `He^(o+),E_(n) = - (21.76 xx 10^(-19) xx z^(2))/(n^(2))`
`E_(2) ("for"He^(o+)) = - (21.76 xx 10^(-19) xx 2^(2))/(3^(2)) = - 9.68 xx 10^(-19) J`
Energy required to remove an e form the third Bohr orbit of `He^(o+)`
`Delta E = 0 - E_(3) = 0 - (-9.68 xx 10^(-19) J)
`= 9.68 xx 10^(-19) J (atom IE is + ve)`
Since `Delta E = (hc)/(Delta E = (6.266 xx 10^(-34) J s xx 3 xx 10^(8) m)/(9.68 xx 10^(-19) J)`
`= 2.05 xx 10^(-7) m = 2050 xx 10^(-10)m = 2050 Å`