Chapter 26.1, Problem 7PP

To determine

The radius of curvature of the beam in the magnetic field.

Answer to Problem 7PP

The radius of curvature of the beam in the magnetic field is $\boxed{r=3.1\times {10}^{-3}\text{m}}$ .

Explanation of Solution

Given:

The strength of the magnetic field is $B=1.5\times {10}^{-2}\text{T}$ .

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

Mass of lithium atoms is $m=7m_{\text{proton}}$ .

  $V_{\text{accl}}=320\text{V}$ .

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) can be simplified as

  $r=\frac{2Vm}{B^{2}q}$  ...... (2)

Substitute $1.5\times {10}^{-2}\text{T}$ for $B$ , $\left(1.6\times {10}^{-19}\text{C}\right)$ for $q$ , $7m_{\text{proton}}$ for $m$ and 320 V for $V$ in equation (2)

  $\begin{array}{l}r=\frac{2\times 320\times 7\times \left(1.67\times {10}^{-27}\right)}{\left(1.5\times {10}^{-2}\right)\times \left(1.6\times {10}^{-19}\right)}\\ \boxed{r=3.1\times {10}^{-3}\text{m}}\end{array}$

Conclusion:

Hence, the radius of curvature of the beam in the magnetic field is $\boxed{r=3.1\times {10}^{-3}\text{m}}$ .