The first ionization energy of hydrogen is `2.179xx10^(-18)` J The second ionization energy of helium atom will be

Options

A. `8.716xx10^(-18)J`
B. `4.358xx10^(-18)J`
C. `5.45xx10^(-17)` J
D. `1.09xx10^(-18) J`

Answer

A

Explanation

According to Bohr's theory , the energy of the electron, E is related to the nuclear charge Z, and the number of electron orbit n, by the equation
`E alpha Z^2//n^2`.
For hydrogen atom, n=1, so , first I.E. of hydrogen
`alpha Z^2=2.179xx10^(-18) J`
second IE of He is the energy involved in removing electron from `He^+` from its first orbit . Here , Z=2 an n=1 .
Hence, `DeltaE(He^+)=4xxDeltaE_"(H)"`
`=4xx2.179xx10^(-18)` J
`=8.716xx10^(-18)` J