bei48482_FM

Hence

Mean lifetime T1.44T 1  2 (12.7)

Tis nearly half again more than T 1  2. The mean lifetime of a radionuclide whose half-

life is 5.00 h is

T1.44T 1  2 (1.44)(5.00 h)7.20 h

Since the activity of a radioactive sample is defined as

R

we see that, from Eq. (12.5),

RN 0 et

This agrees with the activity law of Eq. (12.2) if R 0 N 0 , or, in general, if

Activity RN (12.8)

Example 12.3

Find the activity of 1.00 mg of radon,^222 Rn, whose atomic mass is 222 u.

Solution

The decay constant of radon is

2.11 10 ^6 s^1

The number Nof atoms in 1.00 mg of^222 Rn is

N2.71 1018 atoms

Hence

RN(2.11 10 ^6 s^1 )(2.71 1018 nuclei)

5.72 1012 decays /s5.72 TBq155 Ci

Example 12.4

What will the activity of the above radon sample be exactly one week later?

Solution

The activity of the sample decays according to Eq. (12.2). Since R 0 155 Ci here and

t(2.11 10 ^6 s^1 )(7.00 d)(86,400 sd)1.28

we find that

RR 0 et(155 Ci)e1.2843 Ci

1.00 10 ^6 kg



(222 u)(1.66 10 ^27 kg /u)

0.693



(3.8 d)(86,400 sd)

0.693



T 1  2

dN



dt

T 1  2



0.693

1





Nuclear Transformations 427

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