The half life of U-238 against alpha decay is 4.5 x 10^9 years, find the activity of 1 kg of U-238:
Correct answer: B. 3.34 x 10^-4 Ci
- A. 2.4 x 10^-4 Ci
- B. 3.34 x 10^-4 Ci
- C. 4.34 x 10^-4 Ci
- D. 2.4 x 10^-5 Ci
Explanation
The number of atoms in 1kg of U-238 (N) = moles x Avagadro's constant = (mass/Mr) x Avagadro's constant = (1000/238) x 6.02x1023 = 2.529x1024Half-life = 4.5x109 yearsConverting this into seconds:(4.5x109) x (3.26x107)1.42x1017sA = λNWhere A is activity in Bq and λ is decay constantλ = 0.693/half-lifeA = (0.693/half-life)NA = (0693/1.42x1017)(2.529x1024)A = 1.234x107 BqConverting Bq to Ci1 Ci = 3.7x1010 BqA = 1.234x107/ 3.7x1010A = 3.34 x 10-4 Ci
Report an error
The more specific you are, the faster it gets fixed. A source beats an opinion.
Prefer email? [email protected]
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
1,257 free Nuclear Physics MCQs from Physics, each with the correct answer and an explanation. Unlimited attempts, no account needed.
Discussion
Stuck on an option, or know a faster way to get there? Ask or explain it here.
No comments yet. Be the first to explain this one.
Exams that ask Physics questions like this
Physics is on 16 papers prepared for on TestUstad, and all of them draw the same bank, so this question is worth knowing for every one of them.
More Nuclear Physics questions
What is the unit of decay constant?
Which particle (marked A) is obtained in the following nuclear reaction?4Be9 + 2He4 -> 6C12 + A
In Nuclear Physics the mass defect is referred to _.
Half life of a radiation active element is that period in which half of the atoms
Iodine-131 is used to trace which cancer
Half life of Au-198 is 2.7 days what will be the activity of 1 mg of Au-198