Question

An unstable particle created at a point P moves with a constant speed of 0.998c until it decays at a point Q. If the lifetime of the particle in its rest frame is 632ns, the distance between points P and Q is ______m. (Rounded off to the nearest integer)
(c=3×108 m/s)

Answer 1

Correct Answer: 2992

Time Dilation and Distance Calculation 

Given Data

• Speed of the particle: \( v = 0.998c \)
• Lifetime of the particle in its rest frame: \( \tau = 632 \) ns
• Speed of light: \( c = 3 \times 10^8 \) m/s

Step 1: Calculate the Time Dilation Factor

The time dilation formula is given by:

\[ \tau' = \frac{\tau}{\sqrt{1 - \frac{v^2}{c^2}}} \]

Substituting \( v = 0.998c \) and \( \tau = 632 \) ns:

\[ \tau' = \frac{632ns}{\sqrt{1 - \frac{(0.998c)^2}{c^2}}} \]

\[ \tau' = \frac{632ns}{\sqrt{1 - 0.996004}} \]

\[ \tau' = \frac{632ns}{\sqrt{0.003996}} \]

\[ \tau' = \frac{632ns}{0.06333} \]

\[ \tau' \approx 9984ns \]

Step 2: Calculate the Distance Between Points P and Q

The distance traveled by the particle is given by:

\[ d = v \times \tau' \]

Substituting \( v = 0.998c \) and \( \tau' = 9984 \) ns:

\[ d = 0.998 \times 3 \times 10^8 \text{ m/s} \times 9984 \times 10^{-9} \text{ s} \]

\[ d = 0.998 \times 3 \times 10^8 \times 9984 \times 10^{-9} \]

\[ d \approx 2992.1 \text{ m} \]

Final Answer

The distance between points P and Q is approximately 2992 m.