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2 years ago

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A beam of unstable K mesons, traveling at a speed of 32c, passes through two counters 9.00 m apart. The particles suffer a negli

gible loss of speed and energy in passing through the counters but give electrical pulses that can be counted. It is observed that 1000 counts are recorded in the first counter and 250 in the second. Assuming that this whole decrease is due to the decay of the particles in flight, what is their half-life as measured in their own rest system?

1 answer:

Anna71 [15]2 years ago

5 0

Answer:

8.66 nano seconds

Explanation:

speed of k mesons ( v ) = c $\sqrt{3} /2$ m/s

Distance between counters ( d ) = 9.00 m

number of countable electrical pulses = 1000 counts in first counter and 250 in second counter

time of travel = d / v = 18 / $\sqrt{3} C$ secs

Next write the decay of particles in lab frame

finally calculate the half life of Meson in its own frame

( t 1/2) of meson in its own frame = 8.66 n-secs

attached below is a detailed solution

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2 years ago

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3 years ago

Read 2 more answers

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Nata [24]

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-900 = 0.6428(1305.5) - 4.9 t²

4.9t² = 839.1754 + 900

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t = 18.84 s

Recall again, ut = 1305.5

u = 1305.5/18.84 = 69.3 m/s

7 0

3 years ago