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Chemistry Question on Bohr’s Model for Hydrogen Atom

Ionisation energy of He+He^+ is 19.6×10−18 J19.6 \times 10^{-18}\, J atom−1^{-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

I.E=Z2n2×13.6eV...(i)I.E= \frac{Z^{2}}{n^{2}}\times13.6 eV\quad\quad...\left(i\right)
or I1I2=Z12n12×n22Z22...(ii)\frac{I_{1}}{I_{2}} = \frac{Z^{2}_{1}}{n^{2}_{1}}\times\frac{n^{2}_{2}}{Z^{2}_{2}}\quad \quad ...\left(ii\right)
Given I1=−19.6×10−18,Z1=2,I_{1} = -19.6 \times 10^{-18}, Z_{1} = 2,
n1=1,Z2=3andn2=1n_{1} = 1, Z_{2 }= 3 and n_{2} = 1
Substituting these values in equation (ii)\left(ii\right).
−19.6×10−18I2=41×19-\frac{19.6\times10^{-18}}{I_{2}} = \frac{4}{1}\times\frac{1}{9}
or I2=19.6×10−18×94\quad I_{2}=19.6\times 10^{-18}\times \frac{9}{4}
=−4.41×10−17 J/atom= -4.41 \times 10^{-17}\,J/atom