Question

The radius of second Bohr orbit, in terms of the Bohr radius, ${𝑎}_0$ , in ${\mathrm{Li}}^{2+}$ is:

• A. $\frac{2{\mathrm{a}}_0}{3}$
• B. $\frac{4{\mathrm{a}}_0}{9}$
• C. $\frac{4{\mathrm{a}}_0}{3}$
• D. $\frac{2{\mathrm{a}}_0}{9}$

Answer 1

The correct option is C

$\frac{4{\mathrm{a}}_0}{3}$

Explanation for correct answer:-

Option (C) $\frac{4{\mathrm{a}}_0}{3}$

Step 1: Calculating the radius of the orbit

For the second Bohr orbit, $\mathrm{n}=2$

Atomic number of ${\mathrm{Li}}^{2+}$is $\mathrm{z}=3$

$$\begin{gathered} \therefore {\mathrm{r}}_{\mathrm{n}}=\frac{{\mathrm{n}}^2\times {\mathrm{a}}_0}{\mathrm{z}} \\ \Rightarrow {\mathrm{r}}_{\mathrm{n}}=\frac{2^2\times {\mathrm{a}}_0}{3} \\ {\mathrm{r}}_{\mathrm{n}}=\frac{4{\mathrm{a}}_0}{3} \end{gathered}$$

Therefore, the radius of second Bohr orbit, in terms of the Bohr radius, ${𝑎}_0$ , in ${\mathrm{Li}}^{2+}$ is option (C) $\frac{4{\mathrm{a}}_0}{3}$