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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=127u.

Step by step solution

01

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脳10-3T.
  3. The period of revolution for the iodine ion is T=1.29ms=1.29脳10-3s.
02

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=2蟺mBq

03

Calculate the mass in atomic mass units

The period of revolution for the iodine ion is

T=2蟺mBq

Therefore, the mass of iodine ion is

localid="1662898371084" m=BqT2蟺

m=(45.0脳10-3T)(1.60脳10-19C)脳(1.29脳10-3s)(7脳(2蟺)脳(1.66脳10-27kg/u))

Where n=7rev is the number of revolutions.

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

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Most popular questions from this chapter

An electron with kinetic energy 2.5keVmoving along the positive direction of an xaxis enters a region in which a uniform electric field B鈫�of magnitude 10kV/mis in the negative direction of the yaxis. A uniform magnetic field is to be set up to keep the electron moving along the xaxis, and the direction of B鈫� is to be chosen to minimize the required magnitude of B鈫�. In unit-vector notation, whatB鈫�should be set up?

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