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;
[tex]N(t) = N_0 \left (\dfrac{1}{2} \right )^{\dfrac{t}{t_{1/2}}[/tex]
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
[tex]t_{1/2}[/tex] = The half life of the radioactive substance = 45 days
Substituting the values gives;
[tex]1 = 6.2 \left (\dfrac{1}{2} \right )^{\dfrac{t}{45}[/tex]
[tex]\dfrac{1}{6.2} = \left (\dfrac{1}{2} \right )^{\dfrac{t}{45}[/tex]
[tex]ln\left (\dfrac{1}{6.2} \right ) = {\dfrac{t}{45} \times ln \left (\dfrac{1}{2} \right )[/tex]
[tex]t = 45 \times \dfrac{ln\left (\dfrac{1}{6.2} \right ) }{ln \left (\dfrac{1}{2} \right )} \approx 118.45 \ days[/tex]
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.
