The isotope57 Co decays by electron capture to Fe with a half-life of

1.	The isotope57 Co decays by electron capture to Fe with a half-life of 272 d. The57 Fe nucleus is produced in an excited state, and it almost instantaneously emits gamma rays. (a) Find the mean lifetime and decay constant for57 Co. (b) If the activity of a radiation source Co is 2.0 µCi now, how many57 Co nuclei does the source contain?(c) What will be the activity after one year?(c) What will be the activity after one year?
Answer» Data: T_1/2 = 272d = 272 × 24 × 60 × 60s = 2.35 × 10⁷ s, A₀ = 2.0uCi = 2.0 × 10^-6 × 3.7 × 10¹⁰ = 7.4 × 10⁴ dis/s t = 1 year = 3.156 × 10⁷ s (a) T_1/2 = \(\cfrac{0.693}{\lambda}\) = = 0.693 \(\tau\) ∴ The mean lifetime for ⁵⁷Co = T = \(\cfrac{T_{1/2}}{0.693}\) = \(\cfrac{2.35\times10^7}{0.693}\) = 3391 x 10⁷ s The decay constant for ⁵⁷Co = λ = \(\frac{1}T\) = \(\cfrac{1}{3.391\times10^7s}\) = 2949 × 10^-8 s^-1 (b) A₀ = N₀ A ∴ N₀ = \(\cfrac{A_0}{\lambda}\) = A₀\(\tau\) = (7.4 × 10⁴ )(3.391 × 10⁷) = 2.509 × 10¹² nuclei This is the required number. (c) A(t) = A0e\(^{-\lambda t}\)= 2e^{-(2.949 x 10\(^{-8}\)) (3.156 x 10\(^{7}\))} = 2e^-0.9307 = 2/e^0.9307 Let x = e^0.9307 ∴ Iog e^x = 0.9307 ∴ 2.303 log₁₀x = 0.9307 ∴ log₁₀^_x= \(\cfrac{0.9307}{2.303}\) = 0.4041 ∴ x = antilog 0.4041 = 2.536 ∴ A(t) = \(\cfrac{2}{2.536}\) μCi = 0.7886 μCi

The isotope57 Co decays by electron capture to Fe with a half-life of 272 d. The57 Fe nucleus is produced in an excited state, and it almost instantaneously emits gamma rays. (a) Find the mean lifetime and decay constant for57 Co. (b) If the activity of a radiation source Co is 2.0 µCi now, how many57 Co nuclei does the source contain?(c) What will be the activity after one year?(c) What will be the activity after one year?

Answer»

Data: T_1/2 = 272d = 272 × 24 × 60 × 60s = 2.35 × 10⁷ s,

A₀ = 2.0uCi = 2.0 × 10^-6 × 3.7 × 10¹⁰ = 7.4 × 10⁴ dis/s

t = 1 year = 3.156 × 10⁷ s

(a) T_1/2 = \(\cfrac{0.693}{\lambda}\) = = 0.693 \(\tau\)

∴ The mean lifetime for

⁵⁷Co = T = \(\cfrac{T_{1/2}}{0.693}\) = \(\cfrac{2.35\times10^7}{0.693}\) = 3391 x 10⁷ s

The decay constant for ⁵⁷Co = λ = \(\frac{1}T\)

= \(\cfrac{1}{3.391\times10^7s}\)

= 2949 × 10^-8 s^-1

(b) A₀ = N₀ A

∴ N₀ = \(\cfrac{A_0}{\lambda}\) = A₀\(\tau\)

= (7.4 × 10⁴ )(3.391 × 10⁷)

= 2.509 × 10¹² nuclei

This is the required number.

= 2e^-0.9307 = 2/e^0.9307

Let x = e^0.9307

∴ Iog e^x = 0.9307

∴ 2.303 log₁₀x = 0.9307

∴ log₁₀^_x= \(\cfrac{0.9307}{2.303}\) = 0.4041

∴ x = antilog 0.4041 = 2.536

∴ A(t) = \(\cfrac{2}{2.536}\) μCi = 0.7886 μCi