Question

The half-life of a certain radioactive substance is 45 days. There are 6.2 grams present initially. On what day

Answer 1

Answer:

There will be less than 1 gram of the radioactive substance remaining by the elapsing of 118 days

Step-by-step explanation:

The given parameters are;

The half life of the radioactive substance = 45 days

The mass of the substance initially present = 6.2 grams

The expression for evaluating the half life is given as follows;

$N(t) = N_0 \left (\dfrac{1}{2} \right )^{\dfrac{t}{t_{1/2}}$

Where;

N(t) = The amount of the substance left after a given time period = 1 gram

N₀ = The initial amount of the radioactive substance = 6.2 grams

$t_{1/2}$ = The half life of the radioactive substance = 45 days

Substituting the values gives;

$1 = 6.2 \left (\dfrac{1}{2} \right )^{\dfrac{t}{45}$

$\dfrac{1}{6.2} = \left (\dfrac{1}{2} \right )^{\dfrac{t}{45}$

$ln\left (\dfrac{1}{6.2} \right ) = {\dfrac{t}{45} \times ln \left (\dfrac{1}{2} \right )$

$t = 45 \times \dfrac{ln\left (\dfrac{1}{6.2} \right ) }{ln \left (\dfrac{1}{2} \right )} \approx 118.45 \ days$

The time that it takes for the mass of the radioactive substance to remain 1 g ≈ 118.45 days

Therefore, there will be less than 1 gram of the radioactive substance remaining by the elapsing of 118 days.