Chapter 29.1, Problem 5PP

To determine

To estimate the number of free electrons in the tip of the Washington monument.

Answer to Problem 5PP

  $1.90\times {10}^{26}$

Explanation of Solution

Given data:

Number of free electrons per centimeter cube in aluminum = $1.81\times {10}^{23}$

Atomic mass of aluminum, m = 2835 g

Density of the aluminum, $\rho$ = 2.702 ${\text{g/cm}}^3$

Formula used:

The volume of the tip $(v)=\frac{m}{\rho }$

Calculation:

The volume of the tip $(v)=\frac{m}{\rho }$

  $\begin{array}{l}v=\frac{2835}{2.702}\\ v=1049c{m}^3\end{array}$

Number of free electrons in the tip of the Washington monument,

Free electrons = $1049\times 1.81\times {10}^{23}$

Free electrons = $1.90\times {10}^{26}$

Conclusion:

Hence, the number of free electrons in the tip of the Washington monument is $1.90\times {10}^{26}$ .