Question

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

The mass of each particle is $5\text{u}$ and the particle is Helium.

Step-by-step solution

Identification of given data

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

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

The magnetic field for nuclear particle is $B=1\text{T}$

The radius of curvature for magnetic field is $R=6.28\text{m}$

The atomic mass unit is equal to the mass of one proton particle and its value is role="math" localid="1663156879499" $\mathbf{1}\mathbf{amu}\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{67}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{27}}\mathbf{kg}$.

Determination of mass of nuclear particle

The mass of nuclear particle is given as:

$m=\frac{qBR}{v}$

Substitute all the values in the above equation.

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

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 $5\text{u}$ and the particle is Helium.