Question

Physicist S. A. Goudsmit devised a method for measuring the mass of heavy ions by timing their period of revolution in a known magnetic field. A singly charged ion of iodine makes 7revin a 45.0mTfield in 1.29ms. Calculate its mass in atomic mass units.

Short answer

The mass of the iodine ion in atomic mass unit is $m=127u.$

Step-by-step solution

Given

1. The number of revolutions of the charged iodine ion is n=7rev.
2. The magnetic field magnitude is $B=45.0mT=45.0\times {10}^{-3}T$.
3. The period of revolution for the iodine ion is $T=1.29ms=1.29\times {10}^{-3}s$.

Understanding the concept

By using formula for the period of revolution for the iodine ion, we can find the mass of the iodine ion.

Formula:

The period of revolution for the iodine ion is

$T=\frac{2\pi m}{Bq}$

Calculate the mass in atomic mass units

The period of revolution for the iodine ion is

$T=\frac{2\pi m}{Bq}$

Therefore, the mass of iodine ion is

localid="1662898371084" $m=\frac{BqT}{2\mathrm{\pi }}$

$m=\frac{(45.0\times {10}^{-3}T)(1.60\times {10}^{-19}C)\times (1.29\times {10}^{-3}s)}{(7\times (2\pi )\times (1.66\times {10}^{-27}kg/u\left)\right)}$

Where n=7rev is the number of revolutions.

Hence the mass of the iodine ion in atomic mass unit is m=127u.