Cesium-137 (₅₅Cs¹³⁷) radioactive byproduct from fission nuclear power station. It has a life half-life of 30 years and emits β^- radiations.

1. Calculate the following reaction equation by entering the appropriate number in the boxes
   ₅₅Cs¹³⁷ → _xBa^x + _x β ^x
2. Show that the decay constant of the Cesium-137 is approximately 7×10^-10S^-.
3. Show that the initial activity of 1kg Cesium-137 is approximately 3×10¹⁵Bq.
4. Explain why 1 kg of Cesium-137, although it has an activity of 3×10¹⁵Bq, would be quite safe in a sealed metal box of a thickness 1 cm.

Solution:

(a) Complete Reaction:

₅₅Cs¹³⁷ à ₅₆Ba¹³⁷ + β ^- + v (antineutrino)

(b) Half Life = T_1/2 ⁼ 30 years = 30 × 365 × 24 × 60 × 60 =9.4608×10⁸

We Know,

$\lambda  = \frac{{0.693}}{{{T_{\frac{1}{2}}}}}$

$\Rightarrow \lambda  = \frac{{0.693}}{{9.4608 \times {{10}^8}}}$

$\therefore \lambda  = 7.32 \times {10^{ - 10}} \approx 7 \times {10^{ - 10}}{S^{ - 1}}$

(C) (c) Given,

Mass (m) = 1 kg = 1000gm

We know,

A₀ = N₀λ

Where, N₀ = Number of atoms, A is the activity and λ is decay constant.

So, 

$\begin{array}{l}{N_0}{\rm{ }} = \frac{{Avogadro\,{\rm{Number }} \times {\rm{ Mass}}}}{{Atomic\;{\rm{Mass}}}}\\ = \frac{{\;6.02{\rm{ }} \times {{10}^{23}} \times 1000}}{{137}}\\ = 4.39 \times {10^{24}}\end{array}$

Thus,

Initial Activity (A₀) = N₀ λ= 4.39×10²⁴  ×  7 × 10^-10 =3.07×10¹⁵Bq

(d)1 kg of cesium-137, although it has an activity of 3×10¹⁵ Bq, would be quite safe in a sealed metal box of a thickness 1 cm. This is because the beta particles emitted by cesium-137 have a very low penetration power. They can be stopped by a few centimeters of air or a thin layer of metal. Therefore, the beta particles would not be able to penetrate the sealed metal box and pose a risk to human health.