Chapter 20, Problem 62P

To determine

Distance between lines formed by the singly charged ions and doubly charged ions of each type on the photographic film.

Answer to Problem 62P

Lines formed by the singly charged ions- distance between the lines of carbon isotopes of mass numbers 12 and 13 is $1.11\times {10}^{-3}\text{ m}$ . Distance between the lines of carbon isotopes of mass numbers 13 and 14 is $1.12\times {10}^{-3}\text{ m}$ .

Lines formed by the singly charged ions- distance between the lines of carbon isotopes of mass numbers 12 and 13 is $5.59\times {10}^{-4}\text{ m}$ . Distance between the lines of carbon isotopes of mass numbers 13 and 14 is $5.60\times {10}^{-4}\text{ m}$ .

Explanation of Solution

Given:

The electric field between the electric plates is $2.48\times {10}^4\text{ V/m}$ .

The magnetic fields is $B=B\text{'}\text{ = 0}\text{.68 T}$ .

Atomic mass of an isotope is $1.67\times {10}^{-27}\text{ kg}$ times the atomic number.

Formula used:

Radius of curvature generated of the magnetic flux is

  $r=\frac{mE}{qBB\text{'}}$

Calculation:

Lines formed has twice its radius of curvature. Thus,

  $2r=\frac{2mE}{qBB\text{'}}$

The radius of curvature of the singly charged carbon isotope 12 is

  $\begin{array}{c}2r_{12}=\frac{2m_{12}E}{qBB\text{'}}\\ \text{= }\frac{2\times 12\times (1.67\times {10}^{-27}\text{ kg})(2.48\times {10}^4\text{ V/m})}{(1.6\times {10}^{-19}\text{ C}){(0.68\text{ T})}^2}\\ \text{= }1.344\times {10}^{-2}\text{ m}\end{array}$

The radius of curvature of the singly charged carbon isotope 13 is

  $\begin{array}{c}2r_{13}=\frac{2m_{13}E}{qBB\text{'}}\\ \text{= }\frac{2\times 13\times (1.67\times {10}^{-27}\text{ kg})(2.48\times {10}^4\text{ V/m})}{(1.6\times {10}^{-19}\text{ C}){(0.68\text{ T})}^2}\\ \text{= }1.455\times {10}^{-2}\text{ m}\end{array}$

The radius of curvature of the singly charged carbon isotope 14 is

  $\begin{array}{c}2r_{14}=\frac{2m_{14}E}{qBB\text{'}}\\ \text{= }\frac{2\times 14\times (1.67\times {10}^{-27}\text{ kg})(2.48\times {10}^4\text{ V/m})}{(1.6\times {10}^{-19}\text{ C}){(0.68\text{ T})}^2}\\ \text{= }1.567\times {10}^{-2}\text{ m}\end{array}$

Now, the distance between the field lines of carbon isotopes 12 and 13 is

  $\begin{array}{c}2r_{13}-2r_{12}=1.455\times {10}^{-2}\text{ m }-1.344\times {10}^{-2}\text{ m}\\ =1.11\times {10}^{-3}\text{ m}\end{array}$

The distance between the field lines of carbon isotopes 13 and 14 is

  $\begin{array}{c}2r_{14}-2r_{13}=1.567\times {10}^{-2}\text{ m - }1.455\times {10}^{-2}\text{ m}\\ =1.12\times {10}^{-3}\text{ m}\end{array}$

Lines formed for the doubly charged ions have twice its radius of curvature. Thus,

  $2r=\frac{2mE}{2qBB\text{'}}$

The radius of curvature of the carbon isotope 12 is $\text{6}\text{.718}\times {10}^{-3}\text{ m}$

The radius of curvature of the carbon isotope 13 is $\text{7}\text{.277}\times {10}^{-3}\text{ m}$

The radius of curvature of the carbon isotope 14 is $\text{7}\text{.837}\times {10}^{-3}\text{ m}$

Now, the distance between the field lines of carbon isotopes 12 and 13 is

  $\begin{array}{c}2r_{13}-2r_{12}=7.277\times {10}^{-3}\text{ m }-\text{ 6}\text{.718}\times {10}^{-3}\text{ m}\\ =5.59\times {10}^{-4}\text{ m}\end{array}$

The distance between the field lines of carbon isotopes 13 and 14 is

  $\begin{array}{c}2r_{14}-2r_{13}=7.837\times {10}^{-3}\text{ m }-\text{7}\text{.277}\times {10}^{-3}\text{ m}\\ =5.60\times {10}^{-4}\text{ m}\end{array}$

Conclusion:

Thus, the distance between the lines of carbon isotopes of mass numbers 12 and 13 is $1.1\times {10}^{-3}\text{ m}$ . Distance between the lines of carbon isotopes of mass numbers 13 and 14 is $1.1\times {10}^{-3}\text{ m}$ .

The distance between the lines of carbon isotopes of mass numbers 12 and 13 is $5.59\times {10}^{-4}\text{ m}$ . Distance between the lines of carbon isotopes of mass numbers 13 and 14 is $5.60\times {10}^{-4}\text{ m}$ .