91影视

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

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

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

$K_p=eV$ 鈥�(颈)

Similarly, for the alpha particle, the kinetic energy is

$K_{伪}=q_{伪}V$

$K_{伪}=2eV$ 鈥�(颈颈)

Taking a ratio of this two equations, we can get

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

Similarly calculate kinetic energy of deuteron as

$K_d=q_{d}V$

Charge on the deuteron is $q_d=+e$

$K_d=eV$ 鈥�(颈颈颈)

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

$\frac{K_d}{K_{伪}}=\frac{eV}{2eV}$

$\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 鈥�(iv)

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

Then velocity can be written as

$v=鈭�(2K/m)$

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

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

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

$r鈭�鈭�mK/q$ 鈥�(惫颈)

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

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

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

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

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

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

=$\sqrt{2}脳10cm$

r_d=14cm.

Similarly, for alpha particle,

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

By substituting the value, we can write

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

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

$r_{伪}=\sqrt{\frac{4脳2}{1}}\frac{1}{2}{r}_p$

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