Question

A beam of particle of half-life 2.0X10-8
sec travels in the laboratory with speed 0.96 times the speed of light. How much distance does the beam travel before the flux of the beam falls to ½ times the initial flux?

Answer 1

Given:

The half life of the particle is 2×10⁻⁸

The speed 0.96 times the speed of light.

To find:

The distance of the beam travel before the flux

Solution:

The speed of light is 3×10⁸

The relation of the frame when the half life was increased.

1/$\sqrt{1-0.96^{2} }$

Now the initial flux relation  with the distance is

d=0.96×3×10⁸×2×10⁻⁸×1/$\sqrt{1-0.96^{2} }$

d=5.76×12.77

d=73.552

So the distance of the beam cover is 73.552 m

Answer 2

Thus the distance covered by the beam cover is 20.56 m

Explanation:

We are given that:

• Half life of the particle = 2.0 x 10^-8  sec
• Falling time = 1/2 times the initial flux

Solution:

The speed of light is 3×10⁸

The relation of the frame when the half life was increased.

1/ √ 1 - 0.96^2

Now the initial flux relation with the distance is

d = 0.96×3×10⁸×2×10⁻⁸× 1 /   √ 1 - 0.96^2

d = 1.92 x 1 / √0.0784

d = 5.76 x 1 / 0.28

d = 5.76× 3.57

d = 20.56

Thus the distance covered by the beam cover is 20.56 m