Answer:
0.805 or 0.781 depending on the method used.
Explanation:
<u>Data</u><u> </u><u>provided</u>
half life ie t1/2= 8days.
No= 50g
t=48days
Nt=?
<u>Metho</u><u>d</u><u> </u><u>1</u>
<u>Formul</u><u>a</u><u>:</u><u> </u><u>N</u><u>t=</u><u> </u><u>No</u><u>(</u><u>1</u><u>/</u><u>2</u><u>)</u><u>^</u><u>n</u>
where n= number of half lives.
= t/t1/2 = 48/8
=6.
thus Nt= 50(1/2)^6= 0.781
<u>Metho</u><u>d</u><u> </u><u>2</u>
<u>Formul</u><u>a</u><u>:</u><u> </u><u>Nt</u><u>=</u><u> </u><u>Noe^-</u><u>(</u><u>lambda</u><u>)</u><u>t</u>
where lambda= decay constant
= ln2/ t1/2= ln2/ 8 = 0.086.
thus Nt= 50e^-(0.086×48)= 0.805.