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 $5 u$ and the particle is Helium.

Step by step solution

Identification of given data

The speed of nuclear particles is $v = 0.710 c$

The charge of the nuclear particle is $q = 2 e = 3.2 \times 10^{- 19} C$

The magnetic field for nuclear particle is $B = 1 T$

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

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

Determination of mass of nuclear particle

The mass of nuclear particle is given as:

$m = \frac{q B R}{v}$

Substitute all the values in the above equation.

$m = \frac{(3.2 \times 10^{- 19} C) (1 T) (6.28 m)}{(0.710 c) (\frac{3 \times 10^{8} m / s}{c})} = 9.435 \times 10^{- 27} kg = (9.435 \times 10^{- 27} kg) (\frac{1 u}{1.66 \times 10^{- 27} kg}) \approx 5 u$