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Chapter 37: Q95P (page 1152)

Ionization measurements show that a particular lightweight nuclear particle carries a double charge (= 2e) and is moving with a speed of 0.710c. Its measured radius of curvature in a magnetic field of 1.00 T is 6.28 m. Find the mass of the particle and identify it. (Hints: Lightweight nuclear particles are made up of neutrons (which have no charge) and protons (charge = e), in roughly equal numbers. Take the mass of each such particle to be 1.00 u. (See Problem 53.)

Short Answer

Expert verified

The mass of each particle is 5u and the particle is Helium.

Step by step solution

01

Identification of given data

The speed of nuclear particles is v=0.710c

The charge of the nuclear particle is q=2e=3.2脳10-19C

The magnetic field for nuclear particle is B=1T

The radius of curvature for magnetic field is R=6.28m

The atomic mass unit is equal to the mass of one proton particle and its value is role="math" localid="1663156879499" 1amu=1.67脳10-27kg.

02

Determination of mass of nuclear particle

The mass of nuclear particle is given as:

m=qBRv

Substitute all the values in the above equation.

m=3.2脳10-19C1T6.28m0.710c3脳108m/sc=9.435脳10-27kg=9.435脳10-27kg1u1.66脳10-27kg鈮�5u

The mass of above particle is nearest mass of Helium particle and number of electrons is also 2.

Therefore, the mass of each particle is 5u and the particle is Helium.

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