Question

(c) - 0.85 eV
Ionisation energy of Het is 19.6 x 10-18 J atom-!. The
energy of the first stationary state (n = 1) of Li2+ is
(a) 4.41 x 10-16 J atom- (b)-4.41 x 10-17 J atom-
(c)-2.2 x 10-15 J atom- (d) 8.82 x 1017 J atom-1
to the
${ { { { { { { {1111.25}^{?} }^{?} }^{?} }^{?} }^{?} }^{?} }^{?21 \\ } }^{2}$
​

Answer 1

Huh sorry I don’t know

Answer 2

Explanation:

Since IEHe+IELi2+=(ZHe+ZLi2+)2=49

∴IELi2+=49×IEHe+=9/4×19.6×10−18

                    =4.41×10−17J/atom

Energy  in stationary state =−IE=−4.41×10−17J/atom

Correct Answer is B