Chapter 28: Q71P (page 833)

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

Expert verified

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

Step by step solution

Given

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

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

$T = \frac{2 π m}{B q}$

Calculate the mass in atomic mass units

The period of revolution for the iodine ion is

$T = \frac{2 π m}{B q}$

Therefore, the mass of iodine ion is

localid="1662898371084" $m = \frac{B q T}{2 π}$

$m = \frac{(45.0 \times 10^{- 3} T) (1.60 \times 10^{- 19} C) \times (1.29 \times 10^{- 3} s)}{(7 \times (2 π) \times (1.66 \times 10^{- 27} k g / u))}$