Question

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

Short answer

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

Step-by-step solution

Listing the given quantities

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

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:

$T=2\pi m_e/qB$

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

We have:

$\begin{array}{c}T=\frac{2\pi m_e}{qB}\\ =\frac{2\pi (9.1\times {10}^{-31}\text{\hspace{0.17em}kg})}{(1.6\times {10}^{-19}\text{\hspace{0.17em}C})(3.53\times {10}^{-3}\text{\hspace{0.17em}T})}\\ =1.01\times {10}^{-8}\text{\hspace{0.17em}s}\end{array}$

But, each particle travels only a half-circular path. So,

$\begin{array}{c}t=\frac{T}{2}\\ =\frac{1.01\times {10}^{-8}s}{2}\\ =5.05\times {10}^{-9}s\end{array}$

Therefore, $5.05\times {10}^{-9}s$after of the decay, $e^{-}$and $e^{+}$will collide.