Chapter 26.1, Problem 5PP

To determine

The mass of an oxygen atom.

Answer to Problem 5PP

The mass of an oxygen atom is $\boxed{m=2.7\times {10}^{-26}\text{kg}}$

Explanation of Solution

Given:

The strength of the magnetic field is $B=7.2\times {10}^{-2}\text{T}$ .

Charge of the particle is $q=1.6\times {10}^{-19}\text{C}$ .

Radius is $r=0.085\text{m}$ .

  $V_{\text{accl}}=110\text{V}$ .

Formula used:

The charge to mass ratio of an ion in a mass spectrometer is

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

Here, $q$ is the charge of an ion, $m$ is mass, $V$ is the potential difference, $B$ is the magnetic field strength and $r$ is the radius of the ions circular path.

Calculation:

Equation (1) simplified as

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

Substitute $7.2\times {10}^{-2}\text{T}$ for $B$ , $1.6\times {10}^{-19}\text{C}$ for $q$ , 0.085 m for $r$ and 110 V for $V$ in equation (2)

  $\begin{array}{l}m=\frac{{\left(7.2\times {10}^{-2}\text{T}\right)}^2\times {\left(0.085\text{m}\right)}^2\times \left(1.60\times {10}^{-19}\text{C}\right)}{2\times \left(110\text{V}\right)}\\ \boxed{m=2.7\times {10}^{-26}\text{kg}}\end{array}$

Conclusion:

Hence, the mass of an oxygen atom is $\boxed{m=2.7\times {10}^{-26}\text{kg}}$