One of the many isotopes used in cancer treatment is 79198au, with a half-life of 2.70 d. Determine the mass of this isotope that is required to give an activity of 285 ci.

Respuesta :

The answer is 1.22 mg

Half-life T=2.7 days =2.7 ×24 ×3600

decay constant =0.693 / T=0.693 / 233280

=2.97 * 10^-6

A=300 Ci

=300 ×3.7 ×10 ^10=1110^10

The relation between activity and the number of isotopes is

A=?n

n=3.73×10^18

Mass of the isotope, m=(3.73×10^18/6.023×10^23)198=1.22mg

What is half-life?

  • It is the time at which the concentration of a substance reduces to half of the initial concentration.
  • The concentration decrease to half of the initial value.
  • It has different formulas for different order reactions.
  • The term is commonly used in nuclear physics to describe how quickly unstable atoms undergo radioactive decay or how long stable atoms survive.
  • The term is also generally used to characterize any type of exponential (rarely, non-exponential) decay.
  • For example, the medical sciences refer to the biological half-life of drugs and other chemicals in the human body. The converse of half-life (in exponential growth) is doubling time.

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