A proton, a deuteron (q=+e, m=2.0u), and an alpha particle (q=+2e, m=4.0u) are accelerated through the same potential difference and then enter the same region of uniform magnetic field, moving perpendicular to . What is the ratio of (a) the proton’s kinetic energy K_p to the alpha particle’s kinetic energy K_a and (b) the deuteron’s kinetic energy K_d to K_a? If the radius of the proton’s circular path is 10cm, what is the radius of (c) the deuteron’s path and (d) the alpha particle’s path

For proton

$$\begin{gathered} q_p=+e \\ {m}_p=1u \\ {r}_p=10cm \end{gathered}$$

For deuteron

$$\begin{gathered} q_d=+e \\ {m}_d=2u \end{gathered}$$

For alpha

$$\begin{gathered} q_{\alpha }=2e \\ {m}_{\alpha }=4u \end{gathered}$$

The kinetic energy of particle is the charge times potential difference applied. From that, we can calculate the kinetic energy of a proton and an alpha particle. Also,by calculating the kinetic energy of a deuteron by using the magnetic force, we can calculate the radius of deuteron.

Formula:

$$\begin{gathered} K=qV \\ F=qvB \end{gathered}$$

From the formula, kinetic energy of proton is

$K_p=q_{p}V$

From the given condition, potential difference is the same for proton and alpha particle. Charge on the proton is +e so that

$K_p=eV$ …(i)

Similarly, for the alpha particle, the kinetic energy is

$K_{\alpha }=q_{\alpha }V$

$K_{\alpha }=2eV$ …(ii)

Taking a ratio of this two equations, we can get

$$\begin{gathered} \frac{K_p}{K_{\alpha }}=\frac{eV}{2eV} \\ \frac{K_p}{K_{\alpha }}=0.50 \end{gathered}$$

Similarly calculate kinetic energy of deuteron as

$K_d=q_{d}V$

Charge on the deuteron is $q_d=+e$

$K_d=eV$ …(iii)

From equation (ii) and (iii), we can find the ratio of kinetic energy of deuteron to the kinetic energy ofthealpha particle.

$\frac{K_d}{K_{\alpha }}=\frac{eV}{2eV}$

$\frac{K_d}{K_a}=0.50$$\frac{K_d}{K_a}=0.50$

We know the magnitude of magnetic force is and it should be equal to centripetal force. From that, we can find the radius of trajectory created by the charged particle.

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

$qB=\frac{mv}{r}$

From that, m …(iv)

But the kinetic energy of the particle is $k=\frac{1}{2}m{v}^2$

Then velocity can be written as

$v=\surd (2K/m)$

Put equation (v) in (iv), and we can get

$$\begin{gathered} r=\sqrt{\frac{2K}{m}}\frac{m}{\mathit{q}\mathit{B}} \\ r=\surd \frac{2mK}{qB} \end{gathered}$$

But the magnetic field is uniform, so we can write as

$r\propto \surd mK/q$ …(vi)

From equation (vi), we can find the ratio ofthedeuteron.

$r_d\propto \frac{\surd \left(m_d{K}_d\right)}{q_d}$And $r_d\propto \frac{\surd \left(m_p{K}_p\right)}{q_p}$

role="math" localid="1662729372942" $\frac{r_d}{r_p}=\frac{{\displaystyle \frac{\surd \left(m_d{K}_d\right)}{q_d}}}{{\displaystyle \frac{\surd \left(m_p{K}_p\right)}{q_p}}}$

$r_d=\frac{{\displaystyle \frac{\surd \left(m_d{K}_d\right)}{q_d}}}{{\displaystyle \frac{\surd \left(m_p{K}_p\right)}{q_p}}}{r}_p$

By substituting the value and using equation ratio $\frac{K_d}{K_p}=0.5$,

$r_d=\sqrt{\frac{\left(m_d{K}_d\right)}{\left(m_p{K}_p\right)}}\frac{q_p}{q_d}{r}_p$

Where K_d/K_P=1

$r_d=\sqrt{\frac{2\times \left(K_d\right)}{1\times \left(K_p\right)}}{r}_p$

=$\sqrt{2}\times 10cm$

r_d=14cm.

Similarly, for alpha particle,

$r_{\alpha }=\sqrt{\frac{\left(m_{\alpha }{K}_{\alpha }\right)}{\left(m_p{K}_p\right)}}\frac{q_p}{q_{\alpha }}{r}_p$

By substituting the value, we can write

$r_{\alpha }=\sqrt{\frac{\left(m_{\alpha }{K}_{\alpha }\right)}{\left(m_p{K}_p\right)}}\frac{q_p}{q_{\alpha }}{r}_p$

where,$K_p/K_{\alpha }=1/2,m_{\alpha }=4u,q_{\alpha }=2e,and{m}_p=1u{q}_p=1e$

$r_{\alpha }=\sqrt{\frac{4\times 2}{1}}\frac{1}{2}{r}_p$

$$\begin{gathered} r_{\alpha }=\sqrt{2}{r}_p \\ =\sqrt{2}\times 10cm \\ {r}_{\alpha }=14cm \end{gathered}$$