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3 years ago

g calculate the effectiveness radiation dosage in sieverts for a 79 kg person who is exposed 6.8x10^9

Physics

1 answer:

Elza [17]3 years ago

4 0

Answer:

The answer is " $\bold{dosage = 0.031 rem}$ "

Explanation:

please find the complete question in the attached file.

Given value:

$m = 79\ kg \\\\n = 3.4 \times 10^9 \\\\E = 5.5 \times 10^{-13} \\\\ RBE = 15$

$\to E = n E\\$

$= 3.4 \times 10^9 \times 5.5 \times 10^{-13} \\\\ = 1.87 \times 10^{-3}$

$\to E(absorbed) = 1.87 \times 10^{-3} \times 0.87 = 1.63 \times 10^{-3}$

calculating the radiation absorbed per kg:

$= \frac{1.63 \times 10^{-3}}{79} \\\\ = 2.06 \times 10^{-5} \\\\ = 0.00206 \ rad$

$\to Dosage = 0.00206 \times 15 \\$

$= 0.031 \ \ rem$

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