Chapter 20, Problem 16P

To determine

The radius of the paths for the deuteron and alpha particle in terms of that for the proton.

Answer to Problem 16P

Solution:

$r_D=\sqrt{2}{r}_P$ and $r_{\alpha }=\sqrt{2}{r}_P$

Explanation of Solution

The radius(r)of the trajectory of a charge particle moving with a velocity(v)perpendicular to magnetic field(B)is given by, $r=\frac{mv}{Bq}$

If K is the kinetic energy of the particle then, $mv=\sqrt{2mK}$

If a charge is accelerated by a potential V then, $mv=\sqrt{2mK}=\sqrt{2mqV}$

$\therefore r=\frac{mv}{Bq}=\frac{\sqrt{2mqV}}{Bq}=\frac{1}{B}\sqrt{\frac{2mV}{q}}$

$\therefore r\propto \sqrt{\frac{m}{q}}$

Given:

A proton (mass m_p), a deuteron (m=2m_p,Q=e)and an alpha particle (m=4m_p,Q=2e)are accelerated by the same potential difference V and then enter a uniform magnetic field B.

Formula used:

(1)$\frac{r_D}{r_P}=\sqrt{\frac{m_D}{q_D}\times \frac{q_P}{m_P}}$

(2)$\frac{r_{\alpha }}{r_P}=\sqrt{\frac{m_{\alpha }}{q_{\alpha }}\times \frac{q_P}{m_P}}$

Calculation:

(1)$\frac{r_D}{r_P}=\sqrt{\frac{m_D}{q_D}\times \frac{q_P}{m_P}}=\sqrt{\frac{2m_P}{e}\times \frac{e}{m_P}}=\sqrt{2}$

$r_D=\sqrt{2}{r}_P$

(2)$\frac{r_{\alpha }}{r_P}=\sqrt{\frac{m_{\alpha }}{q_{\alpha }}\times \frac{q_P}{m_P}}=\sqrt{\frac{4m_P}{2e}\times \frac{e}{m_P}}=\sqrt{2}$

$r_{\alpha }=\sqrt{2}{r}_P$