Question

The half life of Radon is 3.80 days what would be its decay constant?

Answer 1

The half life (T) of the isotope Rn-222, which is (confusingly) also called radon, is approximately 3.8 days. This means that half of the activity present at time t will have decayed at t+3.8 days. The decay constant is defined as ln(2)/T.

Answer 2

Answer:

Decay constant $=2,1 \cdot 10^{-6}$

Explanation:

Given: The half life of Radon

To find: Decay constant

Solution:

Decay equation:

$N(t)=N_{0} \cdot e^{-\lambda t}$

$N(t) - the quantity at time t$

$N_{0}=N(0)$ is the initial quantity, i.e. the quantity at time $t=0$.

$\lambda$ - decay constant

Time required for the decaying quantity to fall to one half of its initial value:

$t_{1 / 2}=\frac{\ln 2}{\lambda}=3.80 \text { days }$

Also we need to convert days in seconds.

1 day$=24 hours =24 \cdot 60 \mathrm{~min}utes \\=24 \cdot 60 \cdot 60 \mathrm{sec}onds$

$\lambda=\frac{\ln 2}{t_{1 / 2}}=\\\frac{0,69}{3.80 \cdot 24 \cdot 60 \cdot 60}\\=\frac{0,69}{328320}\\=2,1 \cdot 10^{-6}$

Decay constant $=2,1 \cdot 10^{-6}$