The decay constant of ¹⁹⁷₈₀ Hg (which decays by electron capture to ¹⁹⁷₇₉ Au ) is 1.8 10⁻⁴ s ⁻¹ . How much time is needed to convert 25 % of this mercury isotope into gold?

Step-by-step solution

Given that 25 % of the mercury isotope is converted into gold, the amount of mercury remaining is 75 % of the initial amount. Let the initial amount be N₀ . The remaining amount at time t is N = 0.75 N₀ = 3 4 N₀ . Using the radioactive decay law: N = N₀ e^ - t 3 4 N₀ = N₀ e^ - t e^ t = 4 3 Taking the natural logarithm on both sides: t = ( 4 3 ) = 4 - 3 Substituting the standard values 4 1.386 and 3 1.098 : t 1.386 - 1.098 = 0.288 Given the decay constant = 1.8 10⁻⁴ s ⁻¹ , the time t is: t = 0.288 1.8 10⁻⁴ t = 0.16