The half-life of tritium (31H) is 12.3 years. If 57.0 mg of tritium is released from a nuclear power plant during the course of a mishap, what mass of the nuclide will remain after 49.2 years? mg

3.57 mg

Explanation:

The following data were obtained from the question:

Half life (t½) = 12.3 years

Original amount (N₀) = 57.0 mg

Time (t) = 49.2 years

Amount remaining (N) =?

Next, we shall determine the rate of disintegration. This can be obtained as follow:

Half life (t½) = 12.3 years

Decay constant (K) =.?

K = 0.693 / t½

K = 0.693 / 12.3

K = 0.0563 /year

Finally, we shall determine the amount remaining after 49.2 years as shown below:

Original amount (N₀) = 57.0 mg

Time (t) = 49.2 years

Decay constant (K) = 0.0563 /year

Amount remaining (N) =?

Log (N₀/N) = kt / 2.303

Log (57/N) = (0.0563 × 49.2) / 2.303

Log (57/N) = 2.76996 / 2.303

Log (57/N) = 1.2028

Take the anti log of 1.2028

(57/N) = anti log (1.2028)

57/N = 15.95

Cross multiply

57 = N × 15.95

Divide both side by 15.95

N = 57 / 15.95

N = 3.57 mg

Thus, 3.57 mg of the nuclide will be remaining after 49.2 years.

i don t now

neutrons are in the middle of the atom and the electrons, which are much smaller.