In a nuclear experiment a proton with kinetic energy 1.0 MeV moves in a circular path in a uniform magnetic field.

(a)What energy must an alpha particle ( $\mathit{q}\mathbf{=}\mathbf{+}\mathbf{2}\mathit{e}\mathbf{,}\mathbf{}\mathit{m}\mathbf{}\mathbf{=}\mathbf{}\mathbf{4}\mathbf{.0}\mathbf{}\mathbf{\text{\hspace{0.17em}u}}$)

(b)What energy must a deuteron ($\mathit{q}\mathbf{=}\mathbf{+}\mathit{e}\mathbf{,}\mathbf{}\mathit{m}\mathbf{=}\mathbf{}\mathbf{2}\mathbf{.0}\mathbf{}\mathbf{\text{\hspace{0.17em}u}}$ ) have if they are to circulate in the same circular path?

i) The kinetic energy of the proton is $K_1=1.0MeV$

ii) Mass of alpha particle is $m_2=4.0\text{u}$

iii) The charge on the alpha particle is $q_2=+2e$

iv) Mass of deuteron particle is $m_3=2.0\text{\hspace{0.17em}u}$

v)Charge one deuteron particle is $q_3=+e$

vi) Mass of proton particle is $m_1=1.0\text{\hspace{0.17em}u}$

vii) Charge one proton particle is $q_1=+e$

We use the formula for kinetic energy in terms of mass and velocity, and from this, we find velocity in terms of kinetic energy and mass. Then, we use the concept that magnetic force is equal to centrifugal force to find the radius. Now, the radius is the same, so equate the radius of the proton with the radius of the alpha particle to get the kinetic energy of the alpha particle. We apply the same formulas for the deuteron particle.

Formula:

$KE=0.5m{v}^2$

$F_m=qvB$

$F_c=\frac{m{v}^2}{r}$

We find the velocity in terms of kinetic energy and mass as follows:

$k=0.5\times m\times v^2$

Rearranging this equation for velocity, we get:

$v=\sqrt{\frac{2k}{m}}$

Now the magnetic force is balanced by centrifugal force so:

$\begin{array}{c}{F}_m=F_C\\ qvB=\frac{m{v}^2}{r}\\ r=\frac{mv}{qB}\end{array}$

Now plug the value of$v$as:

$r=\frac{m}{qB}\times \sqrt{\frac{2k}{m}}$

Now radius for proton as:

$r_1=\frac{m_1}{q_{1}B}\times \sqrt{\frac{2K_1}{m_1}}$

Now radius for the alpha particle as:

$r_2=\frac{m_2}{q_{2}B}\times \sqrt{\frac{2K_2}{m_2}}$

And for deuteron particle is :

$r_3=\frac{m_3}{q_{3}B}\times \sqrt{\frac{2K_3}{m_3}}$

It is given that the radius of the circular path is the same, so :

$r_1=r_2=r_3$

For alpha particle:

$\begin{array}{c}\frac{m_1}{q_{1}B}\times \sqrt{\frac{2K_1}{m_1}}=\frac{m_2}{q_{2}B}\times \sqrt{\frac{2K_2}{m_2}}\\ \frac{K_2}{K_1}=\frac{m_1}{m_2}\times \frac{q_2^2}{q_1^2}\end{array}$

Substituting the values in the above expression, and we get,

$\begin{array}{c}\frac{K_2}{1.0}=\frac{1\text{\hspace{0.17em}u}}{4\text{\hspace{0.17em}u}}\times \frac{{\left(2e\right)}^2}{{\left(e\right)}^2}\\ K_2=1.0MeV\end{array}$

The energy of an alpha particle ($q=+2e,m=4.0\text{\hspace{0.17em}u}$) is $1.0MeV$.

Now for the deuteron particle,

$\frac{K_3}{K_1}=\frac{m_1}{m_3}\times \frac{q_3^2}{q_1}$

Substituting the values in the above expression, and we get,

$\begin{array}{c}\frac{K_3}{1.0}=\frac{1\text{\hspace{0.17em}u}}{2\text{\hspace{0.17em}u}}\times \frac{e^2}{e^2}\\ K_3=0.5MeV\end{array}$

The energy of a deuteron ($q=+e,m=2.0\text{\hspace{0.17em}u}$ ) is $0.5MeV$.