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To cause death within hours of exposure to radiation, the dose needs to be very high, 10Gy or higher, while 4-5Gy will kill within 60 days, and less than 1.5-2Gy will not be lethal in the short term. However all doses, no matter how small, carry a finite risk of cancer and other diseases. Patients exposed to radiation between 8 to 30 Gy experience nausea and severe diarrhea within an hour, and they die between 2 days and 2 weeks after exposure. Absorbed doses greater than 30 Gy cause neurological damage
A radioactive isotope is a species of an element with the same and equal chemical element but posses varying mass numbers.
Iridium is a radioactive isotope with a half-life of 74 days.
i.e.
[tex]\mathbf{ T_{1/2} = 74 \ days}[/tex]
Let assume that:
- the size of a radioactive element = N
- at a time = t
where;
- population reduction = dN
- at time = dt
∴
The rate of change can be expressed by the equation:
[tex]\mathbf{\dfrac{dN}{dt}=-\lambda N}[/tex]----- (1)
Here:
λ = decay constant
By integrating equation (1) and making the decay constant the subject of the formula:
The formula for calculating the decay constant can be expressed as:
[tex]\mathbf{\lambda = \dfrac{In{(2)}}{T_{1/2}}}[/tex]
[tex]\mathbf{\lambda = \dfrac{0.693}{T_{1/2}}}[/tex]
[tex]\mathbf{\lambda = \dfrac{0.693}{74}}[/tex]
[tex]\mathbf{\lambda = 0.009365}[/tex]
Using the first-order reaction of chemical kinetics to determine the time required for Iridium to drop below the safety level, we have;
[tex]\mathbf{t = (\dfrac{1}{k}) In (\dfrac{a}{a-x})}[/tex]
where;
a = initial amount of the radioisotope = 1000
a - x = final amount = 1
∴
[tex]\mathbf{t = (\dfrac{1}{0.009365}) In (\dfrac{1000}{1})}[/tex]
t = 737.6 days
Therefore, we can conclude that the amount of time it requires iridium to drop below the safety level is =737.6 days
Learn more about chemical kinetics here:
https://brainly.com/question/13787506?referrer=searchResults
