Question

(a) If the mean proper life-time of muons is 2.2 micro seconds, what average distance
would they travel in vacuum before decaying in the reference frame in which its
velocity is measured as 0.6c
(b) Compare this distance with the distance the muon sees while travelling

Answer 1

Answer:

2.75 microseconds

Explanation

apply time dilation formula and directly u will get

Answer 2

Given:

• Time, ($T_0$) = 2.2×$10^{-6}$ s
• Velocity, V = 0.6c

To Find:

• (a) The average distance'
• (b) Comparision of their distance with the distance of the motion seen while travelling.

Solution:

(a) The average distance is, D = VT → (equation 1) where 'V' is the velocity and 'T' is the time taken, therefore we should first find the value of 't' to find the distance.

Time, T = $\frac{T_0}{(1-\beta ^2)^{1/2}}$ = $\frac{2.2*10^{-6}}{(1-0.62^2)^{1/2} }$

T = $\frac{2.2*10^{-6}}{(0.6156)^{1/2} }$ = $\frac{2.2*10^{-6}}{0.784 }$

T = 2.80 ×$10^{-6}$

Substitute the values of 'T' and 'V' in equation 1.

The average distance,

D = 0.6×3×$10^8$×2.80×$10^{-6}$

D = 5.04×$10^{-2}$ = 504 m

(b) $D_0$ = V$T_0$ = 0.6×3×$10^8$×2.2×$10^{-6}$ = 3.96×$10^{-2}$

$D_0$ = 396 m

∴(a)  The average distance, D = 504 m

 (b) The difference is, D = 504m and $D_0$ = 396m