Question

The half life of radon is 3.80 days. What would be its decay constant?

Answer 1

The half life of radon is 3.80 days. What would be its 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:

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

Also we need to convert days in seconds.

$1\,day = 24\,hours = 24 \cdot 60\,minutes = 24 \cdot 60 \cdot 60\,seconds$

$$\lambda = \frac{\ln 2}{t_{1/2}} = \frac{0.69}{3.80 \cdot 24 \cdot 60 \cdot 60} = \frac{0.69}{328320} = 2.1 \cdot 10^{-6}$$

**Answer:**

Decay constant $= 2.1 \cdot 10^{-6}$