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According to Bohr's theory, the electro...

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+)`

Text Solution

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According to Bohr.s theory
`E_(n)=(21.76xx10^(-19))/(n^(2))J`
The electronic energy of `He^(+)` ion in the nth Bohr orbit `=(21.76xx10^(-19))/(n^(2))xxZ^(2)J`
Where Z=2
Therefore, energy of `He^(+)` in the 3rd Bohr orbit `=(21.76xx10^(-19))/9xx4J`
`DeltAE=E_(alpha)=E_(3)`
`0=[-(21.76xx10^(-19)xx4)/9]`
`=(21.76xx10^(-19)xx4)/9`
`DeltaE=(hc)/(lamda)implieslamda=(hc)/(DeltaE)`
`=(6.625xx10^(-34)xx3xx10^(8)xx9)/(21.76xx10^(-19)xx4)=2.055xx10^(-7)m`
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