1.

The ionisation energy of He^(+) " is " 19.6 xx 10^(-18)J per atom. Calculate the energy of the first stationary state of Li^(2+)

Answer»

Solution :The ionisation energy of `He^(+) " is " 19.6 xx 10^(-18)J` per ATOM
`therefore` energy of the first ORBIT of `He^(+) (Z=2)= 19.6 xx 10^(-18)J`
`therefore` enegy of the first orbit of `H^(+) (Z=1) = (19.6 xx 10^(-18))/(4)J`
`therefore` energy of the first obit of `Li^(2+) (Z=3) = (19.6 xx 10^(-18))/(4) xx 9`
`=4.41 xx 10^(-17)J`


Discussion

No Comment Found

Related InterviewSolutions