Chapter 26.1, Problem 13SSC

To determine

To find: The mass of the oxygen atom.

Answer to Problem 13SSC

The mass of the oxygen atom is $\boxed{m=2.80\times {10}^{-26}\text{kg}}$ .

Explanation of Solution

Given:

Accelerated potential is $V=232\text{V}$ .

Strength of a magnetic field is $B=75\text{mT}$ .

Radius of the circular path is $8.3\text{cm}$

Formula used:

The charge to mass ratio of an ion can be expressed as

  $\frac{q}{m}=\frac{2V}{B^2{r}^2}$  ...... (1)

Here, $q,m,V,B$ and $r$ are the charge of the ion, its mass, potential difference, magnetic field strength and radius of the path.

Calculation:

Rearranging the expression of equation (1) and obtain the expression for mass $\left(m\right)$

  $m=\frac{q{B}^2{r}^2}{2V}$  ...... (2)

Charge of doubly ionized oxygen atom is $2\left(1.6\times {10}^{-19}\text{C}\right)=3.20\times {10}^{-19}\text{C}$

Substitute $3.20\times {10}^{-19}\text{C}$ for $q$ , 0.075 T for $B$ , $8.3\text{cm}$ for $r$ and 232 V for $V$ in equation (2)

  $\begin{array}{l}m=\frac{\left(3.20\times {10}^{-19}\text{C}\right){\left(0.075\text{B}\right)}^2{\left(8.3\text{cm}\right)}^2}{2\times 232\text{V}}\\ m=\frac{\left(3.20\times {10}^{-19}\text{C}\right){\left(0.075\text{T}\right)}^2{\left(0.083\text{m}\right)}^2}{2\times 232\text{V}}\\ \boxed{m=2.80\times {10}^{-26}\text{kg}}\end{array}$

Conclusion:

Hence, the mass of the oxygen atom is $\boxed{m=2.80\times {10}^{-26}\text{kg}}$ .