Question

Consider 2 grams of radon-222 (t_1/2 = 3.82 days). (a) What is its radioactivity initially? (b) How many radon-222 nuclei remain after 30 days? Show set up and calculation.

Answer 1

Activity = dN/dt = No
Where is the decay constant, = 0.693/t_1/2
= 0.693/(3.82 x 24 x 60 x 60)
= 2.0997 x 10^-6 s^-1.

No is the initial number of nuclei,
222 g of radon contains 6.023 x 10²³ atmos
2 g contains, No = (6.023 x 10²³) x (2/222)
= 5.426 x 10²¹
Activity = No
= (2.0997 x 10^-6) x (5.426 x 10²¹)
= 1.139 x 10¹⁶ Bq

b)
Number of decayed radon nuclei, N = No e^(-t)
= (5.426 x 10²¹) x e^{[(-0.693/3.82) x 30]}
= 0.0234 x 10²¹

Number of remaining nuclei = No - N
= (5.426 x 10²¹) - (0.0234 x 10²¹)
= 5.402 x 10²¹