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 required accelerating voltage is 8.95 V.

Step-by-step solution

given

Given

B = 0.240 T
e= 1.60 x 10^-19
r= 12.7 cm,
z= +1
m= 5000g/ mol

Explanation

The accelerating voltage can be calculated by:
$V=\frac{B^2{r}^{2}ez}{2m}.....1$

Where,
B= the magnetic field strength is
r= radius of the curvature
e= electronic charge
m= mass
z= charge and
V= accelerating voltage

Now substitute values in equation (1)

Now substitute Values
B = 0.240 T
e= 1.60 x 10^-19
r= 12.7 cm,
z= +1
m= 5000g/ mol

$$\begin{gathered} V=\frac{(0.240\mathrm{T}{)}^2(12.7\mathrm{cm}{)}^2\left(1.60\times {10}^{-19}\mathrm{C}\right)(+1)}{2(5000\mathrm{g}/\mathrm{mol})} \\ =\frac{{\left(0.240\mathrm{T}\times \frac{1{\mathrm{Vssmm}}^2}{1\mathrm{T}}\right)}^2{\left(12.7\mathrm{cm}\times \frac{1\mathrm{m}}{100\mathrm{cm}}\right)}^2\left(1.60\times {10}^{-19}\mathrm{C}\right)\left(1\right)}{2\left(5000\frac{\mathrm{g}}{\mathrm{mol}}\times \frac{1\mathrm{mol}}{6.02\times {10}^{23}\mathrm{ions}}\times \frac{1\mathrm{kg}}{1\mathrm{g}}\right)} \\ =\frac{{\left(0.2401\mathrm{Vs}/{\mathrm{m}}^2\right)}^2(0.127\mathrm{m}{)}^2\left(1.60\times {10}^{-19}\mathrm{C}\right)}{2\left(8.306\times {10}^{-24}\mathrm{kg}/\text{ion}\right)}\times \left(\frac{1\mathrm{V}}{1(\mathrm{Vs}{)}^2{\mathrm{C}}^2{\mathrm{m}}^2\mathrm{kg}}\right) \\ =8.95\mathrm{V} \end{gathered}$$

So the accelerating voltage required is 8.95 V.