Question

The half-life of ²²²Rn is 3.82 days.

(a) Convert the half-life to seconds.
s

(b) Calculate the decay constant for this isotope.
s^?1

(c) Convert 0.650 ?Ci to the SI unit the becquerel.
Bq

(d) Find the number of ²²²Rn nuclei necessary to produce a sample with an activity of 0.650 ?Ci.
²²²Rn nuclei

(e) Suppose the activity of a certain ²²²Rn sample is 6.20 mCi at a given time. Find the number of half-lives the sample goes through in 40.2 d and the activity at the end of that period. (Enter your answer for the number of half-lives to at least one decimal place.)

	half-lives
	mCi

Answer 1

a)

Given

Half life = 3.82 days

= 3.82 x 24 hrs

= 3.82 x 24 x 60 min

= 3.82 x 24 x 60 xx 60 sec

so

half life = 330048 sec

b)

we know that

decay constant = 0.693 / t1/2

decay constant = 0.693 / 330048

decay constant = 2.09969 x 10-6 s-1

c)

we know that

1 Ci = 3.7 10^10 Bq

so

0.65 Ci = 0.65 x 3.7 x 10^10

so

0.65 Ci = 2.405 x 10^10 Bq

d)

we know that

activity = lamda x N

so

2.405 x 10^10 = 2.09969 x 10-6 x N

N = 1.145 x 10^16

so

1.145 x 10^16 nuclei is required

e)

number of half lives = 40.2 / 3.82

number of half lives = 10.5

we know that

A = Ao e^(-lamda x t )

so

A= 6.20 x e^( - 0.693 x 40.2 / 3.82 )

A= 4.218 x 10-3

so

activity at the end of period is 4.218 x 10-3 mCi