What is the activity of a radioactive sample containing 1.0 x 10^12 nuclei with decay constant lambda = 2.0 x 10^-4 s^-1?
Answer and explanation
Correct answer: A. 2.0 x 10^8 Becquerels (Bq)
- Step 1: Radioactive decay activity: A = lambda N. Step 2. A = (2.0 x 10^-4 s^-1) (1.0 x 10^12) = 2.0 x 10^8 disintegrations per second (Bq). Answer: 2.0 x 10^8 Becquerels (Bq). Why not the others: 5.0 x 10^15 divides N by lambda; 2.0 x 10^16 multiplies by 10^4.
Practise more Atomic, Nuclear and Quantum Physics.
Free questions with an explanation for every one, and full timed ECAT Pakistan mock tests.
More Atomic, Nuclear and Quantum Physics questions
- In Compton scattering, what is the maximum possible wavelength shift (Compton shift delta lambda) when a photon collides with a free station…
- What is the energy required to excite an electron in a hydrogen atom from its ground state (n = 1) to the first excited state (n = 2)?
- According to Heisenberg's uncertainty principle, if the position of an electron is determined with an uncertainty of delta x = 1.0 x 10^-10…
- In a nuclear fusion reaction, four protons fuse to form a helium-4 nucleus with a mass defect of delta m = 0.0287 atomic mass units (u). How…
- What is the velocity of an electron whose de Broglie wavelength is equal to its Compton wavelength (lambda_c = h / (m_e * c))?
- What nuclear particle is emitted when cobalt-60 (Z = 27, A = 60) decays into nickel-60 (Z = 28, A = 60)?