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+)`

Answer

2

Explanation

Let us first calculaye the energy required to remove an electron from the third orbit of the `He^(+)` ion. This energy will be equal to the energy released `(DeltaE)` when an electron will drop from `oo` orbit to third orbit. For `He^(+), Z=2` and for `H^(+), Z=1`, thus for `He^(+)`
`DeltaE=-(21.76xx10^(-19))/(oo^(2))xx2^(2)-((21.76xx10^(-19))/3^(2)xx2^(2))=0+9.67xx10^(-19) J`
Now, `DeltaE=hv=(hc)/lambda`
`:. lambda=(hc)/(DeltaE)=((6.626xx10^(-34))(3xx10^(8)))/(9.67xx10^(-19))=2.055xx10^(-7)` metres.