Question

The radioisotope phosphorus-32 is used in tracers for measuring phosphorus uptake by plants. The half-life of phosphorus-32 is 14.3 days. If you begin with 30.5 mg of this isotope, what mass remains after 27.5 days have passed? Since the decomposition is a radioactive decay reaction, it is first order.

Answer 1

Answer:

The mass remains after 27.5 days is 8.043 mg.

Explanation:

Given that:

Half life = 14.3 days

$t_{1/2}=\frac{\ln2}{k}$

Where, k is rate constant

So,  

$k=\frac{\ln2}{t_{1/2}}$

$k=\frac{\ln2}{14.3}\ days^{-1}$

The rate constant, k = 0.04847 days⁻¹

Time = 27.5 days

Using integrated rate law for first order kinetics as:

$[A_t]=[A_0]e^{-kt}$

Where,  

$[A_t]$ is the concentration at time t

$[A_0]$ is the initial concentration  = 30.5 mg

So,  

$[A_t]=30.5\times e^{-0.04847\times 27.5}\ mg=8.043\ mg$

The mass remains after 27.5 days is 8.043 mg.