Question

17. The ionisation energy of Het is 19.6 x 10-18 J. The 2
energy of first stationary state of Li2+ is:
(A) - 4.9 x 10-28 J (B) -9.8 x 10-18 J
(C) -8.9 x 10-17J (D) – 4.41 x 10-17
​

Answer 1

En = -13.6 x z2/ n2 ev

IE of He+= 19.6 x 10-18 x 2^2/1^2

= 19.6 x10-18 x 4/1

where, Z(atomic no.) = 2

n = 1

for Li+ , IE = ?

Z= 3

n= 1

I.E. = -13.6 x 3^2/1 .........(2)

divide(ii)/(i)

I.E.of Li2+/I.E. of He+

=I.E. of Li+/19.6x 10-18 = 9/4

after solving ,

I.E. of Li2+= 4.41x10-17 j