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The half-life of a radioactive substance is approximately 68.7 days. A sample has an initial count rate of 106 counts per minute. When will the count rate drop to 53 counts per minute? Explain the sig

Physics

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Answers

2 parts
a

68.67 days68.67 \text{ days}

b
94

Step-by-step solution

9 steps

Part a· 4 steps

**Help the people to know the value of half-life and advise them accordingly.

  1. Determine the initial count rate.

    From the table, at Time = 0 days, the initial count rate (N0N_0) is 106 min−1106 \text{ min}^{-1}.

  2. Calculate the count rate after one half-life.

    After one half-life, the count rate will be half of the initial count rate:

    N1/2=N02=106 min−12=53 min−1N_{1/2} = \frac{N_0}{2} = \frac{106 \text{ min}^{-1}}{2} = 53 \text{ min}^{-1}
  3. Estimate the half-life from the table.

    We need to find the time when the count rate is 53 min−153 \text{ min}^{-1}.
    Looking at the table:

    • At 62 days, the count rate is 56 min−156 \text{ min}^{-1}.
    • At 82 days, the count rate is 47 min−147 \text{ min}^{-1}. The count rate of 53 min−153 \text{ min}^{-1} falls between these two points. We can use linear interpolation to estimate the half-life (T1/2T_{1/2}):
    T1/2=62 days+(82 days−62 days)×56 min−1−53 min−156 min−1−47 min−1T_{1/2} = 62 \text{ days} + (82 \text{ days} - 62 \text{ days}) \times \frac{56 \text{ min}^{-1} - 53 \text{ min}^{-1}}{56 \text{ min}^{-1} - 47 \text{ min}^{-1}} T1/2=62 days+20 days×39T_{1/2} = 62 \text{ days} + 20 \text{ days} \times \frac{3}{9} T1/2=62 days+20 days×13T_{1/2} = 62 \text{ days} + 20 \text{ days} \times \frac{1}{3} T1/2=62 days+6.67 daysT_{1/2} = 62 \text{ days} + 6.67 \text{ days} T1/2≈68.67 daysT_{1/2} \approx \boxed{68.67 \text{ days}}
  4. Advise the people based on the half-life.

    The scientists recommended that the environment would be safe for use again if the half-life value is less than 60 days. Since the calculated half-life is approximately 68.67 days, which is greater than 60 days, the environment is not yet safe for use according to the scientific recommendation. The people should be advised to continue to avoid the area for a longer period than initially estimated, as the radioactive materials are decaying slower than the safety threshold.

Part b94

**Sensitize the people about the risk associated with radioactive materials and how they should be handled.

Radioactive materials emit radiation that can be very harmful to living organisms. Exposure to radiation can cause cell damage, leading to serious health issues such as cancer, genetic mutations, birth defects, and acute radiation sickness. To handle radioactive materials safely, it is crucial to follow the principles of time, distance, and shielding. Minimize the time spent near the source, maximize the distance from the source, and use appropriate shielding materials (like lead or concrete) to block radiation. Never touch radioactive materials directly, and always dispose of them according to strict safety protocols established by experts.

c) Help the people to understand the value of number of neutrons of a nuclide Y formed when a radioactive substance X of mass number 208 and atomic number 104 decays by emitting 3 alpha particles, 4 beta particles and gamma rays to form a nuclide Y.

  1. Write down the initial nuclide X.

    The radioactive substance X has a mass number of 208 and an atomic number of 104.

    104208X^{208}_{104}\text{X}
  2. Write down the decay equation.

    The decay involves emitting 3 alpha (α\alpha) particles, 4 beta (β\beta) particles, and gamma (γ\gamma) rays to form nuclide Y.

    • An alpha particle is 24He^{4}_{2}\text{He}.
    • A beta particle is −10e^{0}_{-1}e.
    • A gamma ray is 00γ^{0}_{0}\gamma (it carries no mass or charge). The decay equation is:
    104208X→ZYAYY+3(24He)+4(−10e)+00γ^{208}_{104}\text{X} \rightarrow ^{A_Y}_{Z_Y}\text{Y} + 3 \left(^{4}_{2}\text{He}\right) + 4 \left(^{0}_{-1}e\right) + ^{0}_{0}\gamma
  3. Apply conservation of mass number (A).

    The total mass number must be conserved on both sides of the equation.

    208=AY+(3×4)+(4×0)+0208 = A_Y + (3 \times 4) + (4 \times 0) + 0 208=AY+12+0+0208 = A_Y + 12 + 0 + 0 208=AY+12208 = A_Y + 12 AY=208−12A_Y = 208 - 12 AY=196A_Y = 196
  4. Apply conservation of atomic number (Z).

    The total atomic number must be conserved on both sides of the equation.

    104=ZY+(3×2)+(4×−1)+0104 = Z_Y + (3 \times 2) + (4 \times -1) + 0 104=ZY+6−4+0104 = Z_Y + 6 - 4 + 0 104=ZY+2104 = Z_Y + 2 ZY=104−2Z_Y = 104 - 2 ZY=102Z_Y = 102
  5. Determine the number of neutrons in nuclide Y.

    The number of neutrons is calculated as Mass Number (A) - Atomic Number (Z).

    Number of neutrons in Y=AY−ZY\text{Number of neutrons in Y} = A_Y - Z_Y Number of neutrons in Y=196−102\text{Number of neutrons in Y} = 196 - 102 Number of neutrons in Y=94\text{Number of neutrons in Y} = \boxed{94}
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