Question

226Ra (t1/2 = 1620 y) decays to produce 222Rn (t1/2 = 3.82 d). A sample containing 4.5 mg of 226Ra was purified and left undisturbed for 12 hours. How much 222Rn was present at this time, measured in disintegration rate in mCi (millicuries)?

Answer 1

Radio active decay is a first order reaction.

For first order recation,

half life t_1/2 = 0.693 /k where k is rate constant

k = 0.693/ t_{1/2 --- Eq (1)}

k = 1/t ln { [A]_o/[A]_t} -----Eq (2)

From Eqs (1) and (2),

0.693/ t_1/2 = (1/t) ln {[A]_o/ [A]_t} ------Eq (3)

Given that

half life of 226Ra t_1/2= 1620 y = 1620 x 365 x 24 hrs

time t = 12 hrs

Initial amount of 226Ra ₌ 4.5 mg

Final amount of 226Ra   [A]_t = ?

Substitute all the values in Eq (3),

0.693/ t_1/2 = (1/t) ln {[A]_o/ [A]_t}

[(0.693)/(1620x365x24)] hrs = (1/12 hrs)  ln {4.5 mg/ [A]_t}

ln {4.5 mg/ [A]_t} = [(0.693)/(1620x365x24)]x 12

-  ln {[A]_t/ 4.5 mg } = [(0.693)/(1620x365x24)]x 12

ln {[A]_t/ 4.5 mg } = - [(0.693)/(1620x365x24)]x 12

  [A]_t/ 4.5 mg = e ^{-[(0.693)/(1620x365x24)]x 12}

   [A]_{t =} (4.5 mg) . e^{-[(0.693)/(1620x365x24)]x 12}

⁼ 4.49 mg

[A]_t =4.49 mg

Final amount of 226Ra = 4.49 mg

Given that 226Ra decays to produce 222Rn.

Therefore, 4.49 mg of 222Rn would be prsent .