Question

5. The ionization energy of He ion is 19.6x10-18J atom-?. Calculate the energy of the first excited state of Li+2 ion​

Answer 1

E=hcR(1/n1^2–1/n2^2)Z^2

where E is energy, c and R are constant, Z IS ATOMIC NUMBER

So for a particular transition in He(Z=4), Energy is 19.6x10^-18(say A),

the energy for the same transition in lithium(z=3) would be,

A*3^2/2^2=9A/4

so energy for Li+2 transition will be 9*19.6*10^-18/4=44.1*10^-18J per atom