Answer:
[tex]\large \boxed{\text{4.5 g}}[/tex]
Explanation:
There are eight isotopes of lawrencium, and they are all radioactive,
I will do the calculation for ²⁶²Lr, which has a half-life of 3.6 h.
Let A₀ = the original amount of lawrencium.
The amount remaining after one half-life is ½A₀.
After two half-lives, the amount remaining is ½ ×½A₀ = (½)²A₀.
After three half-lives, the amount remaining is ½ ×(½)²A₀ = (½)³A₀, and so on.
We can write a general formula for the amount remaining:
A =A₀(½)ⁿ
where n is the number of half-lives
[tex]n = \dfrac{t}{t_{\frac{1}{2}}}[/tex]
Data:
A₀ = 5 g
t = 30 min
[tex]t_{\frac{1}{2}} = \text{3.6 h}[/tex]
Calculations:
(a) Convert the half-life to minutes
[tex]t_{\frac{1}{2} }= \text{3.6 h} \times \dfrac{\text{60 min}}{\text{1 h}} = \text{216 min}[/tex]
(b) Calculate n
[tex]n = \dfrac{30}{216} = 0.1389[/tex]
(c) Calculate A
[tex]A = \text{5 g} \times \left (\dfrac{1}{2}\right)^{0.1389} = \text{5 g} \times 0.9082 = \textbf{4.5 g}\\\\\text{The mass of lawrencium remaining after 30 min is $\large \boxed{\textbf{4.5 g}}$}[/tex]
