Question

A radioactive substance decays according to the following function, where yo is the initial amount present, and y is the amount present at time t (in days). y=y_o e^-0.0645t Find the half-life of this substance. Do not round any intermediate computations, and round your answer to the nearest tenth days

Answer 1

Answer:

Therefore 10.75 days is the half life of the given radioactive substance.

Step-by-step explanation:

Given that, a radioactive substance decays function is

$y=y_0 \ e^{-0.0645t}$

where y₀= initial amount of radioactive substance

y= The amount of radioactive substance at time t.

t is in days

Half life is the time required to reduce the amount of the substance to its half of the initial amount.

For half life

$y=\frac{1}{2} y_0$

Putting the value of y in the given function

$\frac12y_0=y_0 \ e^{-0.0645t}$

$\Rightarrow \frac12= \ e^{-0.0645t}$

Taking ln both sides

$\Rightarrow ln( \frac12)= ln(e^{-0.0645t})$

$\Rightarrow ln1-ln2=-0.0645t$     [ $\because lne^a=a,$$ln(\frac ab)= ln \ a- ln \ b$  ]

$\Rightarrow-ln2=-0.0645t$

$\Rightarrow t=\frac{-ln2}{-0.0645}$

      =10.75 days.

Therefore 10.75 days is the half life of the given radioactive substance.