91Ó°ÊÓ

Initial kinetic energy,$K=80 \text{MeV}$

Rest energy of the neutron,$m{c}^2=135 \mathrm{MeV}$

Mean life of the neutron is, $\mathrm{τ}=8.3×{10}^{-17} \text{s}$

The expression for the Lorentz factor is given by,

$\mathrm{γ}=1+\frac{K}{m{c}^2}$

Here role="math" localid="1663156598914" $\mathrm{γ}$is the Lorentz factor, K is the kinetic energy, $m{c}^2$is rest energy of the neutron.

Using the expression of Lorentz factor,

$\mathrm{γ}=1+\frac{K}{m{c}^2}$……. (1)

Substitute 80 MeV forK and 135 MeV for$m{c}^2$ in the above equation,

$\begin{array}{rcl}\mathrm{γ}& =& 1+\frac{80 MeV}{135 MeV}\\ & =& 1+0.59\\ & =& 1.59\end{array}$

The expression for the Lorentz factor in terms of speed is given by,

$\mathrm{γ}=\frac{1}{\sqrt{1-{\left(\frac{v}{c}\right)}^2}}$……. (ii)

Substitute 1.59 for $\mathrm{γ}$and$2.99×{10}^8 \text{m/s}$ for c into the equation (ii)

$\begin{array}{rcl}1.59& =& \frac{1}{\sqrt{1-{\left(\frac{v}{2.99×{10}^8}\right)}^2}}\\ v& =& 2.33×{10}^8 \text{m/s}\end{array}$

Life time pion in the laboratory frame is,

$t=\mathrm{γ}\mathrm{τ}$…… (iii)

Substitute 1.59 for$\mathrm{γ}$ and $8.3×{10}^{-17} \text{s}$for$\mathrm{τ}$ into the equation (iii)

$\begin{array}{rcl}t& =& \left(1.59\right)\left(8.3×{10}^{-17}\right)\\ & =& 1.3×{10}^{-16}s\end{array}$

The expression for the longest possible track for this particle leaves in a bubble chamber is given by,

$x=vt$….. (iv)

Substitute $2.33×{10}^8 \text{m/s}$for v and$1.3×{10}^{-16}\mathrm{s}$ for t into the equation (iv)

$\begin{array}{rcl}x& =& \left(2.33×{10}^8\right)\left(1.3×{10}^{-16}\right)\\ & =& 3.1×{10}^{-8}\mathrm{m}\end{array}$

Therefore the longest possible track for this particle leaves in a bubble chamber is $\begin{array}{rcl}& & 3.1×{10}^{-8}\mathrm{m}\end{array}$.