Considering the Bohr model of hydrogen like atoms, the ratio of the radius of $5^{\text {th }}$ orbit of the electron in $\mathrm{Li}^{2+}$ and $\mathrm{He}^{+}$is

<p>In the Bohr model for a hydrogen-like atom, the radius of the $ n^{\text{th}} $ orbit is given by</p> <p>$r_n = \frac{n^2 a_0}{Z}$</p> <p>where:</p> <p><p>$ n $ is the principal quantum number,</p></p> <p><p>$ a_0 $ is the Bohr radius, and</p></p> <p><p>$ Z $ is the atomic number (the effective nuclear charge).</p></p> <p>For the $ 5^{\text{th}} $ orbit ($ n = 5 $):</p> <p><p>For $ \mathrm{Li}^{2+} $ (where $ Z = 3 $):</p> <p>$r_5(\mathrm{Li}^{2+}) = \frac{5^2 a_0}{3} = \frac{25a_0}{3}$</p></p> <p><p>For $ \mathrm{He}^{+} $ (where $ Z = 2 $):</p> <p>$r_5(\mathrm{He}^{+}) = \frac{5^2 a_0}{2} = \frac{25a_0}{2}$</p></p> <p>Now, the ratio of the radius of the 5th orbit of $ \mathrm{Li}^{2+} $ to that of $ \mathrm{He}^{+} $ is:</p> <p>$$ \frac{r_5(\mathrm{Li}^{2+})}{r_5(\mathrm{He}^{+})} = \frac{\frac{25a_0}{3}}{\frac{25a_0}{2}} = \frac{25a_0}{3} \times \frac{2}{25a_0} = \frac{2}{3} $$</p> <p>Thus, the ratio is $ \frac{2}{3} $.</p> <p>The correct answer is Option B.</p>