Homework SourseCalculate the activity of 1 mg of 226Ra in mCi I got 1 mCi
Calculate the activity of 1 mg of 226Ra in mCi. ( I got 1 mCi) Determine the mass needed to obtain the same amount of activity for the following nuclei: 241 Am, 224 Ra, and 3 H.
Solution
the number of nuclei is,
N = [(1 x 10-3 g) / (226 g /mol)] (6.02 x 1023 atoms / mol)
= 2.66e+18 nuclei
the activity is calculated as follows:
R = N
= [(0.693)/(T1/2)]N
= [(0.693) / (1600 years)(3.1556926e+7 s/year)](2.66e+18)
= 3.65x107 decays / s
= 3.65x107 decays / s[mCi / 3.7x10^7 decays/s]
= 0.986 mCi
= 1 mCi (nearly)
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for 241 Am:
the number of nuclei is,
N = R /
= (3.65x107) /[(0.693)/(T1/2)]
= (3.65x107) /[(0.693)/(432.2 years)]
= (3.65x107) /[(0.693)/(432.2 years)(3.1556926e+7 s/year)]
= 7.19e+17
Thus the number of nuclei is
N = nNA
N = [m/M]NA
Thus, the mass m is,
m = NM/NA
= [7.19e+17][241 g/mol] / (6.02x1023)
= 0.288x10-3 g
= 0.288 mg
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for 224 Ra:
the number of nuclei is,
N = R /
= (3.65x107) /[(0.693)/(T1/2)]
= (3.65x107) /[(0.693)/(3.64 days)]
= (3.65x107) /[(0.693)/(3.64 days)(86400 s/day)]
= 1.656e+13
Thus the number of nuclei is
N = nNA
N = [m/M]NA
Thus, the mass m is,
m = NM/NA
= [1.656e+13][224 g/mol] / (6.02x1023)
= 6.163x10-9 g
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for 3 H :
the number of nuclei is,
N = R /
= (3.65x107) /[(0.693)/(T1/2)]
= (3.65x107) /[(0.693)/(12.35 years)]
= (3.65x107) /[(0.693)/(12.35 years)(3.1556926e+7 s/year)]
= 2.055e+16
Thus the number of nuclei is
N = nNA
N = [m/M]NA
Thus, the mass m is,
m = NM/NA
= [2.055e+16][3 g/mol] / (6.02x1023)
= 1.02x10-7 g
