Question

(2) Pions ("-ons) are subatomic particles with a half-life of 2.6 x 10 seconds as measure by a clock at rest with respect to them. This means that after the passage of this amount of this time as measured by a clock traveling with them, half of them will have decayed. A batch of 50,000 pions with β-092 is created during an experiment at CERN. They are directed along a vacuum line to a detector 50 meters away. How many will arrive at the detector? (5 points) HINT: the two events (in the "stationary" lab frame) are: (i) creation of the pions at x-0, and (ii) their arrival at the detector at x-50 meters away, while they travel at v 0.92c. The number of particles which still survive after N half- lives is 1/2N of the original number.

Answer 1

The time felt by the Pions in there frame will be running slower, (There will be time dilation) Therefore the half-life of the pions in their frame is,

$t' = \frac{t_0}{\sqrt{1-\frac{v^2}{c^2}}}$

Substituting value we get,

$\Rightarrow t_{1/2}' = \frac{2.6 \times 10^{-8}}{\sqrt{1-\frac{(0.92c)^2}{c^2}}} = \frac{2.6 \times 10^{-8}}{\sqrt{1-0.92^2}} = 6.63 \times 10^{-8} \, s$

Now with respect to the Pion frame the distance will remain same as 50 m (There will be Length contraction from the Lab frame of reference).

Therefore the time taken by the Pions to reach detector is,

$t = \frac{50}{0.92c} = 1.81 \times 10^{-7} \, s$

Now we have to calculate the value of N for calculating number of particles remaining,

$N = \frac{t}{t'_{1/2}} = \frac{1.81 \times 10^{-7}}{6.63 \times 10^{-8}} = 2.73 \, \text{half lives}$

Therefore the number of particles reaching the detector is,

$N' = \frac{N_0}{2^N} \Rightarrow N' = \frac{50000}{2^{2.73}} = 7536.29 \simeq 7536$

Therefore 7536 particles will be reaching the detector after travelling 50 meters.