91影视

Consider the formula for the number of copper ions as:

$\mathbf{Number}\text{鈥�}\mathbf{of}\text{鈥�}\mathbf{copper}\text{鈥�}\mathbf{ions}\mathbf{=}\frac{\mathrm{distance}}{a}$

Consider the formula for the density of atoms:

$畏=\frac{density}{atomic\text{鈥�}mass}$

Solve for the mass of the copper:

$\begin{array}{c}m=\left(63.546\text{鈥塽}\right)\left(\frac{1.66脳{10}^{鈭�27}\text{鈥塳驳}}{\text{1鈥塽}}\right)\\ =1.055脳{10}^{鈭�25}\text{鈥塳驳}\end{array}$

Solve for the number of atoms per unit volume as:

$\begin{array}{c}畏=\frac{density}{atomic\text{鈥�}mass}\\ =\frac{8.9脳{10}^3\text{鈥�}\frac{\text{kg}}{{\text{m}}^{\text{3}}}}{1.055脳{10}^{鈭�25}\text{鈥塳驳}}\\ =8.437脳{10}^{28}\text{鈥�}\frac{\text{atoms}}{{\text{m}}^3}\end{array}$

Solve for the relaxation time as:

$\begin{array}{c}蟿=\frac{m}{e^2畏蚁}\\ 蟿=\frac{9.11脳{10}^{鈭�31}\text{鈥塳驳}}{(1.6脳{10}^{鈭�19}\text{鈥塁})(8.437脳{10}^{28}{\text{鈥尘}}^{鈭�3})(4脳{10}^{鈭�9}\text{鈥�}惟鈰�\text{m})}\\ 蟿=1.054脳{10}^{鈭�13}\text{鈥塻}\end{array}$

Determine the speed of the electron as:

$\begin{array}{c}v=\sqrt{\frac{2\left(KE\right)}{\text{m}}}\\ =\sqrt{\frac{2\left(7.0\text{鈥塭痴}\right)(1.6脳{10}^{鈭�19}\text{鈥塉})}{9.11脳{10}^{鈭�31}\text{鈥塳驳}}}\\ =1.568脳{10}^6\text{鈥�}\frac{\text{m}}{\text{s}}\end{array}$

Solve for the distance travelled by the electrons as:

localid="1660041976047" $\begin{array}{c}d=v蟿\\ =\left(1.568脳{10}^6\mathrm{鈥�}\frac{m}{s}\right)(1.054脳{10}^{鈭�13}s)\\ =0.17\mathrm{鈥�}渭m\end{array}$

Consider the inverse of $畏$is the volume per atom and is solved as:

$\begin{array}{c}a=\frac{1}{畏}\\ =2.28脳{10}^{鈭�10}\text{鈥尘}\end{array}$

Solve for the number of ions as:

$\begin{array}{c}n=\frac{\text{distance}}{a}\\ =\frac{1.7脳{10}^{鈭�7}\text{鈥尘}}{2.28脳{10}^{鈭�10}\text{鈥尘}}\\ 鈮�700\end{array}$

Therefore, the distance travelled$0.17\text{鈥�}渭\text{m}$is and the number of ions is 700.