Question

Technetium-99m is an ideal radioisotope for scanning organs because it has a half-life of 6.0h and is a pure gamma emitter. Suppose that 80.0mg were prepared in the technetium generator this morning. How many milligrams of technetium 99m would remain after each of the following intervals?

a. one half-life

b. two half-lives

c. 18 h

d.1.5 days

Short answer

Part (a).$40mg$

Part (b). $20mg$

Part (c). $10mg$

Part (d). $1.25g$

Step-by-step solution

Given Information (Part a)

Half-life is the time expected for an amount to decrease to half of its underlying worth. The term is ordinarily utilized in atomic physical science to portray how rapidly unsteady particles go through radioactive rot or how long stable iotas get by.

Explanation Part (a)

Given,

Initial amount = $80mg$

Calculating the half-life ,

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

After one half-life = localid="1654416625983" $\frac{80mg}{2}=40mg$

Explanation Part (b)

Given,

Initial amount = $80mg$

Calculating the half-life ,

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

After two half-lives = localid="1654416641960" $\frac{80mg}{2^2}=20mg$

Explanation Part (c)

Given,

Initial amount = $80mg$

Calculating the half-life ,

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

After localid="1654416661266" $18h$or three half-lives =localid="1654416654163" $\frac{80mg}{2^3}=10mg$

Explanation Part (d)

Given,

Initial amount = $80mg$

Calculating the half-life,

$1.5$days = localid="1653309928181" $\frac{36}{6}=6$half lives

localid="1654416680168" $\Rightarrow \frac{80}{2^6}=1.25mg$