What isotope is the product of the alpha (*2He) nuclear decay of Radon-222 (222U)? 222 Rn
86
a +
2

ⁿₐX => ²¹⁸₈₄Po

Explanation:

Let ⁿₐX be the isotope.

Thus, the equation can be written as follow:

²²²₈₆Rn —> ⁴₂α + ⁿₐX

Next, we shall determine the value of 'n' and 'a'. This can be obtained as follow:

222 = 4 + n

Collect like terms

222 – 4 = n

218 = n

Thus,

n = 218

86 = 2 + a

Collect like terms

86 – 2 = a

84 = a

Thus,

a = 84

ⁿₐX => ²¹⁸₈₄Po

²²²₈₆Rn —> ⁴₂α + ⁿₐX

²²²₈₆Rn —> ⁴₂α + ²¹⁸₈₄Po

part 1: the half-life of the element is 22 years.

part 2: it will take 132.02 year for the sample to decay from 308.0 g to 4.8125 g.

explanation:

part 1: there are two ways to determine the half life time.

first way: the half life time can be determined directly from its definition as it is the time required for the sample to decay to its half concentration.

from the data the initial concentration was 45.0 g and decayed to its half value (22.5 g) in 22.0 years.

second way: the radio active decay is considered first order reaction so we can use the laws of first order reactions.

k = (1/t) ln (a/a-x), where k is the rate constant of the reaction,

t is the time of the reaction

a is the initial concentration of the reactant

a-x is the remaining concentration at time (t)

a = 45.0 g the initial concentration at t = 0 year.

at t = 22 year, a-x = 22.5 g

then k = (1/22 year) ln (45.0 g/ 22.5 g) = 0.0315 year-1.

the half-life time (t1/2) = ln 2/ k = 0.693/ k =0.693/ 0.0315 = 22 years.

part 2:

using the same law k = (1/t) ln (a/a-x),

using the given data; a = 308.0 g and a-x = 4.8125 g and the calculated k from the part 1 (k = 0.0315 year-1)

t = (1/k) ln (a/a-x) = (1/0.0315) ln (308.0/ 4.8125) = 132.02 years