Question

Calculate the energy of first stationary state of $Li ^{2+}$ if ionisation energy of $He ^{+}$ is $19.6 \times 10^{-18} J$ atom $^{-1}$

• A. $176.4 \times 10^{-18} J\, atom^{-1}$
• B. $4.9 \times 10^{-18} \,J\, atom^{-1}$
• C. $8.7 \times 10^{-18} J\, atom ^{-1}$
• D. $44.1 \times 10^{-18} J\, atom^{-1}$

Answer 1

Answer: (D)

$E_{1}$ i.e. $I.E.$ of $He ^{+}=E_{1}$ of $H \times 2^{2}$

$E_{1}$ i.e. I.E. of $Li ^{2+}=E_{1}$ of $H \times 3^{2}$

$\frac{E_{1} \text { of } L i^{2+}}{E_{1} \text { of } He^{+}}=\frac{9}{4}$

$\Rightarrow E_{1} \text { of } L i^{2+}$

$=\frac{9}{4} \times 19.6 \times 10^{-18}$

$=44.1 \times 10^{-18} J \text { atom }^{-1} $