Question

Calculate the radiation dosage (in grays) for an 79-kg person that is exposed for 6.0 seconds to a 3.0 C source of beta radiation. Assume that 100% of the radiation is absorbed and each beta particle has an energy of 1.6 10-13 J

Answer 1

Answer:

The absorbed energy by the person is $6.74\times10^{-4}\ grays$

Explanation:

Given that,

Mass of person = 79 kg

Time = 6.0 s

Beta radiation = 3.0 C

Energy$E = 1,6\times10^{-13}\ J$

We need to calculate the number of disintegration in 3 ci

$N=3.0\times\dfrac{3.7\times10^{10}\ disintegration/s}{ci}\times3.0\ s$

$N=3.33\times10^{11}\ disintegration$

We need to calculate the total energy of beta particle

$E=3.33\times10^{11}\times1.6\times10^{-13}\ J$

$E=0.05328\ J$

We need to calculate the energy absorbed by the 79 kg of person

$E'=\dfrac{E}{m}$

Put the value into the formula

$E'=\dfrac{0.05328}{79}$

$E'=6.74\times10^{-4}\ J/kg = grays$

Hence, The absorbed energy by the person is $6.74\times10^{-4}\ grays$