Question

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

Short answer

1. The ratio of proton’s kinetic energy K_p to the alpha particle’s kinetic energy K_a is,$\frac{K_p}{K_a}=0.50.$

1. The ratio of deuteron’s kinetic energy K_d to K_a is$\frac{K_d}{K_a}=0.50$

1. Radius of deuteron’s path is $r_d=14cm$
2. Radius of the alpha particle’s path is $r_{\alpha }=14cm$

Step-by-step solution

 Step 1: Identification of given data

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

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

Understanding the concept

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:

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

(a) Determining the ratio of the proton’s kinetic energy Kp to the alpha particle’s kinetic energy Ka .

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

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

(b) Determining the ratio of the deuteron’s kinetic energy Kd to Ka 

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$

(c) Determining the radius of the deuteron’s path

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.

(d) Determining the radius of the alpha particle’s path

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

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

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