An alpha particle can be produced in certain radioactive decays of nuclei and consists of two protons and two neutrons. The particle has a charge of$\mathbf{q}\mathbf{=}\mathbf{\text{+2e}}$ and a mass of $\mathbf{\text{4.00u}}$, where $\mathbf{\text{u}}$is the atomic mass unit, with${\mathbf{\text{1u=1.661×10}}}^{\mathbf{\text{-27}}}$ kg. Suppose an alpha particle travels in a circular path of radius$\mathbf{\text{4.50}}$ cm in a uniform magnetic field with$\mathbf{B}\mathbf{=}\mathbf{\text{1.20T}}$ . Calculate (a) its speed (b) its period of revolution, (c) its kinetic energy, and (d)the potential difference through which it would have to be accelerated to achieve this energy.

The charge on the alpha particle is,$\text{q}=+2\text{e}$

Mass of alpha particle is,$\text{m}=4.0\text{0\hspace{0.17em}u}$

Atomic mass unit is,$1\text{\hspace{0.17em}u}=1.661\times 1{\text{0}}^{\text{-27}}\text{\hspace{0.17em}kg}$

The radius of the circular path is,localid="1663950003398" $\text{r}=4.5\text{0cm}\left(\frac{1m}{100cm}\right)=0.045m$

The magnetic field is,$\text{B}=1.2\text{0T}$

Use the concept of centripetal force and magnetic force to find the speed of the particle. Use the concept of period, kinetic energy, and potential difference. Using equations, find the period of revolution, kinetic energy, and potential difference.

Formulae are as follows:

${\text{F}}_{\text{CP}}=\frac{{\text{mv}}^{\text{2}}}{\text{r}}$

${\text{F}}_{\text{B}}=\text{qvB}$

$\text{T}=\frac{\text{2πr}}{\text{v}}$

$\text{K.E.}=\frac{\text{1}}{\text{2}}{\text{mv}}^{\text{2}}$

$\text{ΔV}=\frac{\left(\text{K.E.}\right)}{\text{q}}$

Where F_B is a magnetic force, B is the magnetic field, v is velocity, m is mass, q is the charge on particle, K.E is kinetic energy, F_CP is the centripetal force, r is the radius, and T is time period.

Speed of the particle:

Alpha particle is moving in a circular path, so the magnetic force provides the centripetal force.

$F_{CP}=F_B$

$\frac{{\text{mv}}^{\text{2}}}{\text{r}}=\text{qvB}$

rearrange it for v;

$\text{v}=\frac{\text{qBr}}{\text{m}}$

Plugging the values,

$\begin{array}{c}v=\frac{(2\times 1.6\times {10}^{-19}C)\left(1.20T\right)\left(0.045m\right)}{(4.00\times 1.661\times {10}^{-27}kg)}\\ =2.60\times {10}^{6}m/s\end{array}$

Hence, the speed of the particle is $2.60\times {10}^{6}m/s$$2.60\times {10}^{6}m/s$

Period of revolution of the particle:

Using the equation of period for circular path,

$\text{T}=\frac{\text{2πr}}{\text{v}}$

$\begin{array}{c}\text{T}=\frac{2\left(3.14\right)\left(0.045m\right)}{2.60\times {10}^{6}m/s}\\ =1.09\times {10}^{-7}s\end{array}$

Hence, the period of revolution of the particle is $1.09\times {10}^{-7}s$

The kinetic energy of the particle:

$\begin{array}{c}K.E.=\frac{1}{2}(4.00\times 1.661\times {10}^{-27}kg){(2.60\times {10}^{6}m)}^2\\ =2.245\times {10}^{-14}J\end{array}$

Hence, the kinetic energy of the particle is $2.245\times {10}^{-14}J$

Potential difference:

Using the equation of potential difference,

$\text{ΔV}=\frac{\left(\text{K.E.}\right)}{\text{q}}$

convert the kinetic energy in eV,

localid="1663950023710" $\begin{array}{c}K.E.=\text{2.245}\times {\text{10}}^{\text{-14}}\text{J}\times \left(\frac{\text{1\hspace{0.17em}eV}}{\text{1.6}\times {\text{10}}^{\text{-19}}\text{\hspace{0.17em}J}}\right)\\ =1.40\times {10}^{5}eV\end{array}$

Therefore,

$\begin{array}{c}\text{ΔV}=\frac{\text{1.40}\times {\text{10}}^{\text{5}}\text{eV}}{\text{2e}}\\ =7.00\times {10}^{\text{4}}\text{\hspace{0.17em}V}\end{array}$

Hence, the potential difference is$7.00\times {10}^{\text{4}}\text{\hspace{0.17em}V}$

Therefore, by using the concept of centripetal force and magnetic force the speed of the particle, kinetic energy, potential difference, and period of revolution can be found.