Chapter 28: Q31P (page 831)

A particular type of fundamental particle decays by transforming into an electron $e^{-}$ and a positron $e^{+}$ . Suppose the decaying particle is at rest in a uniform magnetic field of magnitude3.53mT and the $e^{-}$ and $e^{+}$ move away from the decay point in paths lying in a plane perpendicular to $\vec{B}$ . How long after the decay do the $e^{-}$ and $e^{+}$ collide?

Short Answer

Expert verified

$e^{-}$ and $e^{+}$ will collide after $5 .05 \times 10^{- 9} s$ following the decay.

Step by step solution

Listing the given quantities

Magnetic field $B = 3.53 m T = 3.53 \times 10^{- 3} T$ .

Understanding the concept of the period of motion

We need to use the formula of the period of motion of charged particle into a magnetic field to find the period of the particle. As each particle travels only a half-circular path, dividing this period by 2, we will get the required time after which andwill collide.

Formula:

$T = 2 π m_{e} / q B$

Calculation of theafter the decay when e-and e+ collide

We have:

$\begin{matrix} T = \frac{2 π m_{e}}{q B} \\ = \frac{2 π (9.1 \times 10^{- 31} kg)}{(1.6 \times 10^{- 19} C) (3.53 \times 10^{- 3} T)} \\ = 1.01 \times 10^{- 8} s \end{matrix}$