The radius of fifth orbit of the $\mathrm{Li}^{++}$ is __________ $\times 10^{-12} \mathrm{~m}$.

Take: radius of hydrogen atom $= 0.51\,\mathop A\limits^o$

<p>The formula to calculate the radius of an orbit for a hydrogen-like atom/ion is:</p> <p>$r_n = r_0 \frac{n^2}{Z}$</p> <p>where:</p> <ul> <li>$r_n$ is the radius of the nth orbit,</li> <li>$n$ is the principal quantum number (the orbit number),</li> <li>$r_0$ is the Bohr radius (radius of the first Bohr orbit in the hydrogen atom), and</li> <li>$Z$ is the atomic number (the number of protons in the nucleus).</li> </ul> <p>We&#39;re dealing with a Li²⁺ ion and we&#39;re interested in the fifth orbit ($n = 5$), and given that $r_0$ is 0.51 Å and $Z$ for Li is 3, we can substitute these values into the formula:</p> <p>$r_5 = 0.51 \times \frac{25}{3} \text{ Å} = 4.25 \text{ Å}$</p> <p>which is $4.25 \times 10^{-10}$ m, or equivalently $425 \times 10^{-12}$ m when converted to meters.</p>