ionisation energy of He is `19.6 xx 10^(-18) j atom^(-1 ) ` The energy of the first stationary state (n=1) of `Li^(+)` is

Options

A. `4.14 xx 10^(16) atom^(-1)`
B. `4.14 xx 10^(-17) J ` atom
C. `2.2 xx 10^(15) J ` atom
D. `8.82 xx 10^(17) J` atom

Answer

B

Explanation

ionisation energy for `He^(+)`
`E_(1) (He^(+)) = -0 19.6 xx10^(18) J atom^(-1)`
where `E_(n) =` ionization energy of `He^(+)`
K = the energy of the first stationary state
`Z=2 `
`k= (-19.6 xx 10^(18))/(4 ) = 4.9 xx 10^(18)`
`E_(1) (L_(1)%^(2)) = (4.9 xx 10^(18) 3^(2))/( 1^(2))` `44.1 xx 10^(18)`
`4.41 xx 10^(17 j atom^(1)`