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

Using the formula for the radius of the nth orbit: \[ r_n = k \cdot \frac{n^2}{Z} \] For Li$^{2+}$ (\textit{Z} = 3, \textit{n} = 2): \[ r_2 = k \cdot \frac{2^2}{3} = \frac{4k}{3} \] For Be$^{3+}$ (\textit{Z} = 4, \textit{n} = 3): \[ r_3 = k \cdot \frac{3^2}{4} = \frac{9k}{4} \] The ratio of radii: \[ \frac{r_3}{r_2} = \frac{\frac{9k}{4}}{\frac{4k}{3}} = \frac{27}{16} \] Thus, the radius of the 3$^{rd}$ orbit of Be$^{3+}$ is $\frac{27}{16}x$.