A sample of a radioactive isotope contains $N_0=1.0\times 10^6$ nuclei. After 15 days only $2.5\times 10^5$ nuclei remain. What is the half‑life of the isotope? (Assume exponential decay).

Step-by-step solution

1. Write the decay law: $N=N_0{\left(\frac{1}{2}\right)}^{t/T_{1/2}}$.
2. Insert the known values: $2.5\times 10^5=1.0\times 10^6{\left(\frac{1}{2}\right)}^{15/T_{1/2}}$.
3. Divide both sides by $N_0$ to obtain $0.25={\left(\frac{1}{2}\right)}^{15/T_{1/2}}$.
4. Recognise that $0.25=(1/2{)}^2$, therefore $\frac{15}{T_{1/2}}=2$.
5. Solve for the half‑life: $T_{1/2}=\frac{15}{2}=7.5\text{days}$.

Concept tested: Half‑life calculation using decay law

Published 29 August 2026 · How this solution was prepared and checked

Related questions

• 1 The fission of a ^235U nucleus after capturing a thermal neutron can be represented as ^235U + ^1n → ^141Ba + ^92Kr + 3 ^1n. Given the atomic masses: m(^235U) = 235.0439 u, m(^1n) = 1.008665 u, m(^141Ba) = 140.9144 u, m(^92Kr) = 91.9262 u. Calculate the total energy released in this fission reaction. (1 u = 931.5 MeV/c^2).
• 2 The atomic mass of a helium‑4 nucleus is 4.002603 u, the mass of a proton is 1.007276 u and the mass of a neutron is 1.008665 u. Calculate the binding energy per nucleon of the helium‑4 nucleus. (1 u = 931.5 MeV/c^2).
• 3 Explain why, for the same kinetic energy, alpha particles travel a shorter distance in air than beta particles.
• 4 A compound microscope uses an objective lens of focal length 4.0 mm and an eyepiece of focal length 25.0 mm. The distance between the lenses (tube length) is 160 mm. Calculate (i) the magnifying power of the objective, (ii) the angular magnification of the eyepiece (assume the final image is formed at the near point 25 cm), and (iii) the overall magnification of the microscope.
• 5 Two thin lenses are placed in contact. A convex lens of power +5.0 D and a concave lens of power -2.0 D are used. Find the focal length of the combination in centimeters.