Final answer:
To find the thickness of lead required to reduce gamma ray intensity to 0.100%, you calculate the logarithm of the desired intensity divided by the logarithm of the absorption rate. Multiplying the result by the initial layer thickness provides the total thickness needed. However, given answer options do not match the calculated result, which suggests reevaluation is needed.
The correct answer is None of all.
Step-by-step explanation:
The problem at hand is one related to the exponential attenuation of gamma rays (γ-rays) as they pass through shielding material. In this case, we are provided with the starting information that a 1.50-cm-thick piece of lead absorbs 90% of γ-rays. This indicates that 10% of the rays can penetrate this thickness of lead. Using the principle that each subsequent layer of lead of the same thickness will absorb the same percentage of γ-rays that enter it, we can calculate the total thickness of lead required to reduce the γ-ray intensity to 0.100% of its original.
To find the number of times n that we need to apply the 90% absorption to reach 0.100%, we use the formula:
Final Intensity = Initial Intensity × (1 - Absorption Rate)^n
Letting the final intensity be 0.100% (0.001 of the original intensity), the initial intensity to be 1 (or 100%), and the absorption rate to be 0.90 (90%), we get:
0.001 = 1 × (0.90)^n
Taking the logarithm of both sides:
log(0.001) = n × log(0.90)
Solving for n gives us:
n = log(0.001) / log(0.90) ≈ 6.578
Since 1.50 cm of lead corresponds to one layer, the total thickness required will be 1.50 cm × n:
Total Thickness = 1.50 cm × 6.578 ≈ 9.87 cm
As 9.87 cm is not one of the options provided, we must check our calculation or our approach.