Question

The radius of the fourth orbit of single electron species is 2.116 A^o. Calculate the energy of the orbit.

• A. -27.2 eV
• B. -13.6 eV
• C. -40.8 eV
• D. -54.4 eV

Answer 1

The correct option is B

-13.6 eV

The explanation for the correct option:

1. According to Bohr, the radius (r_n) of single-electron species for n^th orbit is

${\mathrm{r}}_{\mathrm{n}}=\frac{{\mathrm{n}}^2}{\mathrm{Z}}\times 0.529\stackrel{\circ }{\mathrm{A}}$

2. Now, r_n = 2.116 A^o, n =4 as the radius of the third orbit is given, thus, atomic number (Z) is calculated as-

$$\begin{gathered} Z=\frac{{\mathrm{n}}^2}{{\mathrm{r}}_{\mathrm{n}}}\times 0.529\stackrel{\circ }{\mathrm{A}} \\ \mathrm{Z}=\frac{4^2}{2.116\mathrm{A}°}\times 0.529\stackrel{\circ }{\mathrm{A}} \\ \mathrm{Z}=\frac{16\times 0.529}{2.116}=4 \end{gathered}$$

3. Hence, the atomic number, Z = 4

4. According to Bohr's model, the energy of the n^th orbit is-

${\mathrm{E}}_{\mathrm{n}}=-13.6\times \frac{{\mathrm{Z}}^2}{{\mathrm{n}}^2}\mathrm{eV}$

5. Now, Atomic number, Z = 4 and n = 4, thus the energy of the 4th orbit is-

${\mathrm{E}}_{\mathrm{n}}=-13.6\times \frac{4^2}{4^2}\mathrm{eV}=-13.6\mathrm{eV}$

6. Hence, the energy of the fourth orbit is -13.6 eV (option b).

The correct option is b).