Question

The half-life of a radioactive isotope is 2.5 min. If there are initially 48 Times 10^32 atoms of this isotope, (a) the number of atoms of this isotope remaining after 15 min is: 12 Times 10^32 7.5 Times 10^31 3 Times 10^32 6 Times 10^4 3 Times 10^2 (b) what is the decay constant? (c) What is the initial activity?

Answer 1

Number of atoms, N_t = N_oe^{-(ln2)t/t_1/2}

So, N₁₅ = (48 * 10³²) * e^{-(ln2 * 15 / 2.5)} = 7.5 * 10³¹ atoms

Decay constant, λ = ln2 / t_1/2 = ln2 / 2.5 = 0.28 min^-1

Initial activity, A_o = λN_o = 0.28 * (7.5 * 10³¹) = 2.1 * 10³¹ decays/min