Respuesta :
Given:-
- Half-life of Caesium-137 = 30.3 years
- Initial amount of Caesium-137,[tex]\sf N_o[/tex]= 60 g
To Find :-
- Amount of caesium left over 90.9 years
Solution:-
The amount of substance left after n half cycle of caesium is given by -
[tex]\green{ \underline { \boxed{ \sf{N_t= \frac{N_o}{2^n}}}}}[/tex]
where
- [tex]\sf N_t = Amount \:left \:after \: n \:half \: cycles[/tex]
- [tex]\sf N_o=Initial \:Amount \:of \: radioactive \: element \:[/tex]
- [tex]\sf n= Number \: of \:half \:cycle [/tex]
[tex]\green{ \underline { \boxed{ \sf{Number \: of \: half \:cycle = \frac{Given\:Time \: period}{Period \: of \: half \:cycle}}}}}[/tex]
[tex]\begin{gathered}\\\implies\quad \sf Number \: of \: half \:cycle = \frac{90.9}{30.3}\\\end{gathered} [/tex]
[tex]\begin{gathered}\\\implies\quad \sf Number \: of \: half \:cycle = 3 \\\end{gathered} [/tex]
Now,
[tex]\begin{gathered}\\\implies\quad \sf N_t= \frac{N_o}{2^n} \\\end{gathered} [/tex]
[tex]\begin{gathered}\\\implies\quad \sf N_t= \frac{60}{2^3} \\\end{gathered} [/tex]
[tex]\begin{gathered}\\\implies\quad \sf N_t= \frac{\cancel{60}}{\cancel{8}} \\\end{gathered} [/tex]
[tex]\begin{gathered}\\\implies\quad \sf N_t= 12.5 g \\\end{gathered} [/tex]
Therefore, 12.5 gram of cesium would be left over 90.9 years.
