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

Explanation

` E_(3H) =-(21. 76 xx 10^(-19))/3^2 =- 2 . 41 xx 10^(-19)` joule
` E_( 3He^+) = Z^2 xx E_(3H) =- 2 . 41 xx 10^(-19) xx 4`
`=- 9. 64 xx 10 ^(-19)` joule
Now if electron is to be remove from (II) orbit energy eqvivalent to ` 9. 64 xx 10 ^(-19)` must be provide .
Therefore
` E_3 = (hc)/( lambda)`
or ` lambda= (hc) /(E_3) = ( 6.625 xx 10^(34) xx 3. 0 xx10^8)/(9.64 xx 10^(-19))`
` =2. 061 xx 10^(-7) m = 2061 xx 10^(-10) m = 2061Å`