Chapter 28, Problem 26P

To determine

The binding energy of the third electron in lithium using the Bohr Theory.

Answer to Problem 26P

The answer is $\text{-3}\text{.4}\text{eV}$

Explanation of Solution

Given info: Number of electron: $\text{n = 2}$

Formula used:

Energy levels of $\text{H}$ atom are given by ${\text{E}}_{\text{n}}\text{=-}\frac{\text{13}\text{.6}}{{\text{n}}^{\text{2}}}\text{eV}$

Where, $\text{n}$ is the principal quantum number

Calculation:

The third electron in lithium is in the $\text{2s}$ sub shell, which is outside the more tightly bound filled $\text{1s}$ shell.

This makes it appear as if there is a “nucleus” with a net charge of $\text{+1e}$ .

The energy of the hydrogen atom is ${\text{E}}_{\text{n}}\text{=-}\frac{\text{13}\text{.6}}{{\text{n}}^{\text{2}}}\text{eV}\mathrm{...}\left(\text{1}\right)$

Substitute the values in equation $\left(\text{1}\right)$

  ${\text{E}}_{\text{2}}\text{=}\text{-}\frac{\text{13}\text{.6}\text{eV}}{{\text{2}}^{\text{2}}}\text{= -3}\text{.4}\text{eV}$

Conclusion:

The required value is $\text{-3}\text{.4}\text{eV}$ .