NTA Abhyas JEE Main2020ChemistryChemical KineticsPractice
1   g of Au 79 198 t 1 2 =   65   h decays by Beta emission and produce stable mercury. How much mercury will be present after 260 h.
Options
- A0.93 g
- B0.85 g
- C1 g
- D0.79 g
Correct answer
A. 0.93 g
Step-by-step solution
A u 19 198 → H g 80 108 + θ - 1 0 t 1 2 = 65 h T = 260 h T = t 1 2 × ′ n ′ Number of Lay live (n) = 260 65 = 4 Amount of undecayed A u = N 0 2 n = 1 2 4 = 1 16 g Amount of decayed Au = 1 - 1 16 = 15 16 = 0 .93 g