Question

The radius of the $2^{\text {nd }}$ orbit of $Li ^{2+}$ is $x$. The expected radius of the $3^{\text {rd }}$ orbit of $Be ^{3+}$ is

• A. $\frac{16}{27} \pi$
• B. $\frac{4}{9} x$
• C. $\frac{9}{4} x$
• D. $\frac{27}{16} x$

Answer 1

The Correct Option is D

To solve this problem, we need to understand the concept of the Bohr model of the hydrogen atom and its extension to other hydrogen-like ions. In the Bohr model, the radius of the electron's orbit (for hydrogen-like ions) is given by the formula:

\(r_n = \frac{n^2 h^2}{4 \pi^2 m k Z e^2}\)

Where:

• \(r_n\) is the radius of the n-th orbit.
• \(n\) is the principal quantum number.
• \(h\) is Planck's constant.
• \(m\) is the electron mass.
• \(k\) is the Coulomb's constant.
• \(Z\) is the atomic number of the element.
• \(e\) is the charge of the electron.

The radius of the orbit depends on \(n^2\) and is inversely proportional to Z, i.e., \(r \propto \frac{n^2}{Z}\).

For \(Li^{2+}\), the atomic number \(Z = 3\), and for \(Be^{3+}\), the atomic number \(Z = 4\).

Given:

• The radius of the 2nd orbit of \(Li^{2+}\) is \(x\).
• We need to find the radius of the 3rd orbit of \(Be^{3+}\).

Using the Formula:

For \(Li^{2+}\), using \(n=2\),

\(r_{Li} = \frac{2^2 \cdot a_0}{3}\) (where \(a_0\) is the Bohr radius)

For \(Be^{3+}\), using \(n=3\),

\(r_{Be} = \frac{3^2 \cdot a_0}{4}\)

Ratio of Radii:

To find the relation, we calculate the ratio:

\(\frac{r_{Be}}{r_{Li}} = \frac{\left(\frac{3^2}{4}\right) a_0}{\left(\frac{2^2}{3}\right) a_0} = \frac{9}{4} \cdot \frac{3}{4} = \frac{27}{16}\)

Result:

Thus, the expected radius of the 3rd orbit of \(Be^{3+}\) relative to the 2nd orbit of \(Li^{2+}\) is \(\frac{27}{16} x\).

The correct answer is: \(\frac{27}{16} x\).