Chapter 28, Problem 46GP

To determine

The average distance from the nucleus for an electron in the innermost orbit of uranium atom and energy required to remove this innermost electron.

Answer to Problem 46GP

The average is distance is $\text{5}\text{.8}\text{×}{\text{10}}^{\text{-13}}\text{m}$

The binding energy is $\text{-}\text{1}\text{.15}{\text{×10}}^{\text{5}}\text{eV}$

Explanation of Solution

Given info: For uranium: $\text{Z}\text{=}\text{92}$

Number: $\text{n}\text{=1}$

Formula used:

Bohr formula, $\text{r}\propto \frac{\text{1}}{\text{Z}}$

Where, $\text{r}$ is radius,

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:

From the Bohr formula for the radius, $\text{r}\propto \frac{\text{1}}{\text{Z}}$

So, ${\text{r}}_{\text{uranium}}\text{= }\frac{{\text{r}}_{\text{1}}}{\text{Z}}\text{ = }\frac{\left(\text{0}\text{.529 }\text{× }{\text{10}}^{\text{-10}}\text{m}\right)}{\text{92}}\text{=}\text{5}\text{.8}\text{×}{\text{10}}^{\text{-13}}\text{m}$

Note that the innermost electron would “see” a nucleus with charge $\text{Ze}$ . So, use the energy of hydrogen atom:

  ${\text{E}}_{\text{1}}\text{=-}\frac{\left(\text{13}\text{.6 eV}\right){\text{Z}}^{\text{2}}}{{\text{n}}^{\text{2}}}\text{=}\text{-}\frac{\left(\text{13}\text{.6}\text{eV}\right){\left(\text{92}\right)}^{\text{2}}}{{\text{1}}^{\text{2}}}\text{=-}\text{1}\text{.15}{\text{×10}}^{\text{5}}\text{eV}$

Hence, the binding energy is $\text{-}\text{1}\text{.15}{\text{×10}}^{\text{5}}\text{eV}$

Conclusion:

All the required values are calculated.