Question

A proton, a deuteron ($\mathbf{q}\mathbf{=}\mathbf{+}\mathbf{e}\mathbf{,}\mathbf{}\mathbf{m}\mathbf{}\mathbf{=}\mathbf{}\mathbf{2}\mathbf{.}\mathbf{0}\mathbf{}\mathbf{u}$), and an alpha particle ($\mathbf{q}\mathbf{=}\mathbf{+}\mathbf{2}\mathbf{e}\mathbf{,}\mathbf{}\mathbf{m}\mathbf{}\mathbf{=}\mathbf{}\mathbf{4}\mathbf{.}\mathbf{0}\mathbf{}\mathbf{u}$) all having the same kinetic energy enter a region of uniform magnetic field, moving perpendicular to it. What is the ratio of (a) the radius r_d of the deuteron path to the radius r_pof the proton path and (b) the radius $r_{\alpha }$of the alpha particle path to r_p?

Short answer

a) Ratio of the radius r_dof the deuteron path to the radius r_p of the proton path,$\frac{r_d}{r_p}=1.4$

b) Ratio of the radius $r_{\alpha }$of the alpha particle path to the radius r_pof the proton path,$\frac{r_{\alpha }}{r_p}=1.0$

Step-by-step solution

Identification of given data

1. Mass of proton,$m_p=1u$
2. Charge of proton,$q_p=+e$
3. Mass of deuteron,$m_d=2u$
4. Charge of deuteron,$q_d=+e$
5. Mass of $\alpha$particle, $m_{\propto }=4u$
6. Charge of $\alpha$particle,$q_{\propto }=+2e$

 Step 2: Understanding the concept

For a charge circulating in a magnetic field, the centripetal force is provided by magnetic force. By using this concept, we can find the radius of the circular path of each particle and then the required ratios.

Formula:

1. Magnetic force,$F_B=qvB$
2. Centripetal force,$F=\left(m{v}^2\right)/r$

Step 3: (a) Determining the ratio of the radius rd of the deuteron path to the radius rp of the proton path  

For a charge circulating in a magnetic field, centripetal force is provided by the magnetic force.

Magnetic force

$F_B=qvB$ …(i)

Centripetal force

$F=\left(m{v}^2\right)/r$ …(ii)

Equating equation (i) and (ii),

$qvB=\left(m{v}^2\right)/r$

Rearranging the equation for radius r$r=\frac{mv}{qB}$

…(iii)

Kinetic energy of the circulation particle is given by

$K=\frac{1}{2}m{v}^2$

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

Using in equation (iii), we get

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

$r=\frac{\sqrt{m}}{q}\times \frac{\sqrt{2K}}{B}$ …(iv)

It is given that the magnetic force and kinetic energy of each particle is the same. Applying equation (iv) to deuteron and proton, we get

$r_d=\frac{\sqrt{m_d}}{q_d}\times \frac{\sqrt{2K}}{B}$

$r_p=\frac{\sqrt{m_p}}{q_p}\times \frac{\sqrt{2K}}{B}$

Taking ratio r_d of r_p,

$\frac{r_d}{r_p}=\frac{\sqrt{m_d}}{q_d}\times \frac{q_p}{\sqrt{m_p}}$

Using all the given values, we get

$\frac{r_d}{r_p}=\frac{\sqrt{2}}{+e}\times \frac{+e}{\sqrt{1}}$

$\frac{r_d}{r_p}=1.414$

Ratio of the radius r_d of the deuteron path to the radius r_pof the proton path:

$\frac{r_d}{r_p}=1.414~1.4$

(b) Determining the ratio of the radius rα of the alpha particle path to rp   

It is given that magnetic force and kinetic energy of each particle is the same. Applying equation (iv) to$\alpha$particle and proton, we get

$r_{\alpha }=\frac{\sqrt{m_{\alpha }}}{q_{\alpha }}\times \frac{\sqrt{2K}}{B}$

$r_p=\frac{\sqrt{m_p}}{q_p}\times \frac{\sqrt{2K}}{B}$

Taking ratio of $r_{\alpha }and{r}_p$:

$\frac{r_{\alpha }}{r_p}=\frac{\sqrt{m_{\alpha }}}{q_{\alpha }}\times \frac{q_p}{\sqrt{m_p}}$

Using all the given values, we get

$\frac{r_d}{r_p}=\frac{\sqrt{4}}{+2e}\times \frac{+e}{\sqrt{1}}$

Ratio of the radius $r_{\alpha }$of the alpha particle path to the radius r_p. of the proton path:

$\frac{r_{\alpha }}{r_p}=1.0$