Question

Calculate the accelerating voltage that would be required to direct singly charged ions of mass $5000$through an instrument that is identical to the one described in Example 20-4.

Short answer

The solution involves the substitution of values in the given formula, $\frac{m}{z}=\frac{B^2{r}^{2}e}{2V}$.

Therefore, the accelerating voltage required to direct singly charged ions of mass 5000 is 8.96 V.

Step-by-step solution

The formula

Accelerating voltage can be calculated by -

$\frac{m}{z}=\frac{B^2{r}^{2}e}{2V}$

Where -

Mass of ion$=m$

Magnetic field strength$=B$

Radius of curvature of the magnetic sector $=r$

Charge of the ion$=z$

Electronic charge $=e=\left(1.60\times {10}^{-19}\mathrm{C}\right)$

Voltage$=V$

Understanding the problem

As per given information in the problem -

Mass of charged ions$=5000\mathrm{g}/\mathrm{mol}$

Magnetic field strength$=B={\left(0.240\mathrm{Vs}/{\mathrm{m}}^2\right)}^2$

Radius of curvature of the magnetic sector$=r=(0.127\mathrm{m}{)}^2$

Charge of the ion$=z=1$

Electronic charge$=e=\left(1.60\times {10}^{-19}\mathrm{C}\right)$

Voltage difference $=?$

First we have to convert mass of charged ions into the $\mathrm{kg}/$ion -

role="math" localid="1645528882756" $$\begin{gathered} m=5000\frac{\mathrm{g}}{\mathrm{mol}}\times \frac{1\mathrm{mol}}{6.02\times {10}^{23}\text{ions}}\times \frac{1\mathrm{kg}}{{10}^3\mathrm{g}} \\ =0.83\times {10}^{-23}\frac{\mathrm{kg}}{\text{ion}} \end{gathered}$$

Calculation

By substituting all values in the equation -

$$\begin{gathered} \frac{m}{z}=\frac{B^2{r}^{2}e}{2V} \\ V=\frac{B^2{r}^{2}ez}{2m} \\ =\frac{{\left(0.240\mathrm{Vs}/{\mathrm{m}}^2\right)}^2(0.127\mathrm{m}{)}^2\times 1.60\times {10}^{-19}\mathrm{C}/\text{ion}\times 1}{2\times 0.83\times {10}^{-23}\mathrm{kg}/\text{ion}} \\ =8.96\mathrm{V} \end{gathered}$$

Accelerating voltage required to direct singly charged ions of mass $5000$isrole="math" localid="1645528973211" $8.96\mathrm{V}$.