Answer:
Option B
Explanation:
$I.E = \frac{Z^{2}}{n^{2}}\times13.6eV$ ....(i)
$\frac{I_{1}}{I_{2}} = \frac{Z_{1}^{2}}{n_{1}^{2}}\times\frac{n_2^2}{Z_2^2}$ ...(ii)
Given I1 = -19.6 x 10-18 Z1 = 2, n1= 1, Z2 = 3 and n2 = 1
Substituting these values in equation (ii).
$-\frac{19.6\times10^{-18}}{I_{2}} = \frac{4}{1}\times\frac{1}{9}$
Or I2 = -$19.6\times10^{-18}\times\frac{9}{4}$ = $-4.4\times10^{-17}Jatom^{-1}$
