The half-life of Au-198 is 2.7 days. What will be the activity of 1 mg of Au-198?
Correct answer: C. 240 Ci
- A. 120 Ci
- B. 200 Ci
- C. 240 Ci
- D. 280 Ci
Explanation
half life T= 2.7 dayshalf life T= 2.7 days 24hour3600secondshalf life T= 233280 smass m= 1 mg= 10-3gAtomic mass= 198 g/molactivity A =?activity A =λNWhere λ is decay constantIt is found by given formula The relationship between the half-life, T1/2, and the decay constant is given byT1/2 = 0.693/λ.λ = 0.693T.λ = 0.693233280sλ =2.9 10-6s-1N is number of atoms given by;N=[ mass/atomic mass ]Avogadros number`N=[ 10-3g/198g/mol ]6.022 x 10^(23)``N= 3.03 x 10^18Now, we will find the activity Aactivity A =λNactivity A =[2.9 10-6s-1][3.03 x 10^18]activity A = 8.8x 10^12 disintegrations/second or 8.8x 10^12 BqFor Curie, 1 Ci= 3.7x 10^10 BqHence,A= 8.8x 10^12 / 3.7x 10^10 A= 240 Ci
Last updated
About Nuclear Physics
Atomic nuclei contain protons and neutrons, with nuclear size, mass defect and binding energy describing their stability. The work covers alpha, beta and gamma decay, decay equations, half-life and decay rate, along with radiation safety and medical applications such as imaging, radiotherapy and tracers.
Practise Nuclear Physics
1,254 free Nuclear Physics MCQs from Physics, each with the correct answer and an explanation. Unlimited attempts, no account needed.
Exams that ask Physics questions like this
Physics is on 13 papers prepared for on TestUstad, and all of them draw the same bank, so this question is worth knowing for every one of them.