Solveeit Logo
Login

Question

Chemistry Question on Structure of atom

Ionisation energy of He+He^+ is 19.6×10−18 J atom−119.6 \times 10^{-18}\, J\, atom^{-1}. The energy of the first stationary state (n=1)(n = 1) of Li2+Li^{2+} is

A

4.41×10−16 J atom−14.41 \times 10^{-16}\, J\, atom^{-1}

B

−4.41×10−17 J atom−1-4.41 \times 10^{-17}\, J\, atom^{-1}

C

−2.2×10−15 J atom−1-2.2 \times 10^{-15}\, J\, atom^{-1}

D

8.82×10−17 J atom−18.82 \times 10^{-17}\, J\, atom^{-1}

Answer

−4.41×10−17 J atom−1-4.41 \times 10^{-17}\, J\, atom^{-1}

Explanation

Solution

The correct option is(B): −4.41×10−17 J atom−1-4.41 \times 10^{-17}\, J\, atom^{-1}.

IEHe+=13.6 ZHe+2[112−1∞2]=13.6ZHe+2IE_{_{He^{+}}} = 13.6\,Z^{2}_{_{He^{+}}}\left[\frac{1}{1^{2}}-\frac{1}{\infty^{2}}\right] = 13.6Z^{2}_{_{He^{+}}} where (ZHe+=2)\left(Z_{_{He^{+}}} = 2\right) Hence 13.6×ZHe+2=19.6×10−18 J atom−1.13.6\times Z^{2}_{_{He^{+}}} = 19.6 \times 10^{-18}\,J\, atom^{-1} . (E1)Li+2=13.6 ZLi+22×112=−13.6 ZHe+2×[ZLi+22ZHe+2]\left(E_{1}\right)_{_{Li^{+2}}} = 13.6\,Z^{2}_{_{Li^{+2}}} \times \frac{1}{1^{2}} = - 13.6\,Z^{2}_{_{He^{+}}}\times\left[\frac{Z^{2}_{_{Li^{+2}}}}{Z^{2}_{_{He^{+}}}}\right] =−19.6×10−18×94=−4.41×10−17 J/atom= - 19.6 \times 10^{-18}\times \frac{9}{4} = -4.41 \times 10^{-17}\, J/ atom