Question

Consider the 𝛼-decay ⁹⁰𝑇ℎ ²³²→⁸⁸𝑅𝑎 ²²⁸ . In an experiment with one gram of ⁹⁰𝑇ℎ ²³², the average count rate (integrated over the entire volume) measured by the 𝛼-detector is 3000 counts s ^-1 . If the half life of ⁹⁰𝑇ℎ ²³² is given as 4.4 × 10¹⁷ s, then the efficiency of the 𝛼-detector is ______(rounded off to two decimal places).
Given: Avogadro’s number = 6.023 × 1023 mol⁻¹

Answer 1

Correct Answer: 0.72 - 0.74

To calculate the efficiency of the α-detector, we need to determine the number of α-particles emitted per second by the sample and compare it with the measured count rate.

Step-by-Step Solution: 

1. Calculate the number of atoms in 1 gram of \(90Th^{232}\):

Given the molar mass of \( 90Th^{232} \) is approximately 232 g/mol, and Avogadro's number is \( 6.023 \times 10^{23} \) atoms/mol, the number of atoms in 1 gram of \( 90Th^{232} \) is:

\[ \text{Number of atoms} = \frac{1 \, \text{g}}{232 \, \text{g/mol}} \times 6.023 \times 10^{23} \, \text{atoms/mol} \] \[ \text{Number of atoms} = 2.595 \times 10^{21} \, \text{atoms} \]

1. Calculate the decay constant \( \lambda \) of \( 90Th^{232} \):

The half-life of \( 90Th^{232} \) is given as \( 4.4 \times 10^{17} \, \text{s} \). The decay constant \( \lambda \) is related to the half-life by the equation:

\[ \lambda = \frac{\ln(2)}{T_{1/2}} = \frac{0.693}{4.4 \times 10^{17} \, \text{s}} = 1.57 \times 10^{-18} \, \text{s}^{-1} \]

1. Calculate the activity \( A \) of the sample:

The activity \( A \) of the sample is given by:

\[ A = \lambda \times \text{Number of atoms} = 1.57 \times 10^{-18} \, \text{s}^{-1} \times 2.595 \times 10^{21} \, \text{atoms} \] \[ A = 4.07 \times 10^{3} \, \text{decays/second} = 4070 \, \text{Bq} \] (Note that \( 1 \, \text{Bq} = 1 \, \text{decay/second} \)).

1. Calculate the efficiency of the α-detector:

The α-detector detects 3000 counts per second. The efficiency \( \eta \) is the ratio of the measured count rate to the actual decay rate:

\[ \eta = \frac{\text{Measured count rate}}{\text{Activity}} = \frac{3000 \, \text{counts/second}}{4070 \, \text{decays/second}} \] \[ \eta = 0.737 \approx 0.74 \]

Final Answer:

The efficiency of the α-detector is approximately \( 0.74 \). Therefore, the answer is 0.72 to 0.74.