Respuesta :
Answer :
(a) The mass that remains after 't' years will be, [tex]\frac{70}{2^{(\frac{t}{30})}}mg[/tex]
(b) The sample remains after 110 years will be, 5.52 mg
(c) The time taken will be, 55.0 years.
Explanation :
Formula used :
[tex]a=\frac{a_o}{2^n}[/tex] ............(1)
where,
a = amount of reactant left after n-half lives
[tex]a_o[/tex] = Initial amount of the reactant
n = number of half lives
And as we know that,
[tex]n=\frac{t}{t_{1/2}}[/tex] ..........(2)
where,
t = time
[tex]t_{1/2}[/tex] = half-life
Now equating the value of 'n' from (2) to (1), we get:
[tex]a=\frac{a_o}{2^{(\frac{t}{t_{1/2}})}}[/tex] ...........(3)
(a) Now we have to calculate the mass that remains after 't' years.
As we are given that,
[tex]a_o[/tex] = 70 mg
[tex]t_{1/2}[/tex] = 30 years
Now put all the given values in formula (3), we get:
[tex]a=\frac{70}{2^{(\frac{t}{30})}}[/tex]
(b) Now we have top calculate the sample remains after 110 years.
As we are given that,
t = 110 years
Now put all the given values in formula (3), we get:
[tex]a=\frac{70}{2^{(\frac{110}{30})}}[/tex]
[tex]a=5.52mg[/tex]
(c) Now we have to calculate the time when sample 1 mg remains.
As we are given that,
a = 1 mg
Now put all the given values in formula (3), we get:
[tex]1=\frac{70}{2^{(\frac{t}{30})}}[/tex]
[tex]t=55.0years[/tex]
