Question

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 1

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

∴ log_10^x =  \(\cfrac{0.9307}{2.303}\) = 0.4041

∴ x = antilog 0.4041 = 2.536

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