Answered

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After 24.0 days 2.00 milligrams of an original 128.0. Milligram sample remain what is the half life of the sample

Sagot :

Rascoidn

Answer:

4 days

either multiply 128 by .5 until you get to 2 counting each time or use 2 formulas ln(n2/n1)=-k(t2-t1) to get k then input k into ln(2)=k*t1/2

n2 is final amount and n1 is beginning and t is either time elapsed as in the first formula or the actual half life that is t1/2

Explanation:

Cricetusidn

The half life of the sample will be "4 days".

According to the question,

Let,

  • The number of half life = n

→  [tex]\frac{N}{N_o} = (\frac{1}{2} )^n[/tex]

By substituting the values, we get

→  [tex]\frac{2}{128} = (\frac{1}{2} )^n[/tex]

→    [tex]\frac{1}{64} = (\frac{1}{2} )^n[/tex]

→ [tex](\frac{1}{2} )^6= (\frac{1}{2} )^n[/tex]

→     [tex]n=6[/tex]

So,

It takes 6 half life just to reduce from 128 mg - 2mg which is also "24 days".

→ [tex]6 \ half \ life = 24 \ days[/tex]

Hence,

→ [tex]Half \ life = \frac{24}{6}[/tex]

                   [tex]= 4 \ days[/tex]

Thus the answer above is right.

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