Question

An electron is moving in the 3rd orbit of ${\mathrm{Li}}^{+2}$ and its separation energy is y. The separation energy of an electron moving in the 2nd orbit of ${\mathrm{He}}^{+}$ is:

• A. $\mathrm{y}$
• B. $\frac{\mathrm{y}}{9}$
• C. $-\frac{\mathrm{y}}{9}$
• D. $\frac{4\mathrm{y}}{9}$

Answer 1

The correct option is A

$\mathrm{y}$

Explanation for the correct answer:-

Option (A) $\mathrm{y}$

Step 1: Calculating separation energy of ${\mathrm{Li}}^{2+}$

We know that separation energy ${\mathrm{E}}_{\mathrm{n}}=13.6\times \frac{{\mathrm{Z}}^2}{{\mathrm{n}}^2}$, where Z= Atomic number, n= the shell number or ${\mathrm{n}}^{\mathrm{th}}$ shell.

It is given to us that the separation energy of ${\mathrm{Li}}^{2+}$ is $\mathrm{y}$

This means that, ${\mathrm{E}}_{\infty }-{\mathrm{E}}_3=\mathrm{y}$

Now,

${\mathrm{E}}_{\infty }=\frac{-13.6}{\infty }\times {\left(3\right)}^2\Rightarrow {\mathrm{E}}_{\infty }=0$

And,

$$\begin{gathered} {\mathrm{E}}_3=\frac{-13.6}{{\left(3\right)}^2}\times {\left(3\right)}^2 \\ \Rightarrow {\mathrm{E}}_3=-13.6 \end{gathered}$$

$\therefore$$$\begin{gathered} {\mathrm{E}}_{\infty }-{\mathrm{E}}_3=\mathrm{y} \\ \Rightarrow \mathrm{y}=0-(-13.6) \\ \Rightarrow \mathrm{y}=13.6...........\left(\mathrm{i}\right) \end{gathered}$$

Step 2: Calculating separation energy of ${\mathrm{He}}^{2+}$

Therefore, the separation energy of ${\mathrm{He}}^{2+}$is:

$$\begin{gathered} {\mathrm{E}}_{\infty }-{\mathrm{E}}_2=\left[\frac{-13.6}{{(\infty )}^2}\times {\left(2\right)}^2\right]-\left[\frac{-13.6}{{\left(2\right)}^2}\times {\left(2\right)}^2\right] \\ =0-(-13.6) \\ =13.6 \\ =\mathrm{y}[\mathrm{From}\mathrm{eqn}(\mathrm{i}\left)\right] \end{gathered}$$

Therefore, option (A) $\mathrm{y}$ is the correct answer.