decay in the discharging process to half of its maximum (V = Vo/2). This time is the half-life: t1= t1/2. 20.82 ms, 3.977v T1= 1.4 m/s Measure the time it takes for the voltage to decay to one-quarter (V = Vo/4 = (Vo/2)2) of its maximum. This is two half-lives: t2 = 2t1/2. Then divide this time by two to find the half-life (t1/2 = t2/2) 20.2-24.00 = 3.2 T2 = 3.2/2 = 1.6m/sMeasure the time it takes for the voltage to grow to one-eighth (V = Vo/8 = (Vo/2)3) of its This is three half-lives, t3 = 3t1/2. Then divide this time by three to find the half-life, t1/2 = t3/3. 20.8 – 22.28= 1.48m/s T3= 1.48/3 = 0.49 m/sTake the average of the three measured values of the half-life. Estimate the precision of the measurement and state it as {half-life ± precision} 1.4 + 1.6 + 0.49 / 3 = 1.16 half-life +- Precision Estimate the time constant from the average half-life.

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Principles of Physics: A Calculus-Based Text

5th Edition

ISBN:9781133104261

Author:Raymond A. Serway, John W. Jewett

Publisher:Raymond A. Serway, John W. Jewett

Chapter30: Nuclear Physics

Section: Chapter Questions

Problem 37P

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decay in the discharging process to half of its maximum (V = V_o/2). This time is the half-life: t1= t_1/2.

20.82 ms, 3.977v

T1= 1.4 m/s

Measure the time it takes for the voltage to decay to one-quarter (V = V_o/4 = (V_o/2)²) of its maximum. This is two half-lives: t2 = 2t_1/2. Then divide this time by two to find the half-life (t_1/2 ₌ t2/2)

20.2-24.00 = 3.2

T2 = 3.2/2 = 1.6m/s

Measure the time it takes for the voltage to grow to one-eighth (V = V_o/8 = (V_o/2)³) of its This is three half-lives, t3 = 3t_1/2. Then divide this time by three to find the half-life, t_1/2 ₌ t3/3.

20.8 – 22.28= 1.48m/s

T3= 1.48/3 = 0.49 m/s

Take the average of the three measured values of the half-life. Estimate the precision of the measurement and state it as {half-life ± precision}

1.4 + 1.6 + 0.49 / 3 = 1.16 half-life +- Precision

Estimate the time constant from the average half-life.

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