Explanation:
Given that,
The disintegration constant of the nuclide, [tex]\lambda=0.0178\ h^{-1}[/tex]
(a) The half life of this nuclide is given by :
[tex]t_{1/2}=\dfrac{ln(2)}{\lambda}[/tex]
[tex]t_{1/2}=\dfrac{ln(2)}{0.0178}[/tex]
[tex]t_{1/2}=38.94\ h[/tex]
(b) The decay equation of any radioactive nuclide is given by :
[tex]N=N_oe^{-\lambda t}[/tex]
[tex]\dfrac{N}{N_o}=e^{-\lambda t}[/tex]
Number of remaining sample in 4.44 half lives is :
[tex]t_{1/2}=4.44\times 38.94[/tex]
[tex]t_{1/2}=172.89\ h^{-1}[/tex]
So, [tex]\dfrac{N}{N_o}=e^{-0.0178\times 172.89}[/tex]
[tex]\dfrac{N}{N_o}=0.046[/tex]
(c) Number of remaining sample in 14.6 days is :
[tex]t_{1/2}=14.6\times 24[/tex]
[tex]t_{1/2}=350.4\ h^{-1}[/tex]
So, [tex]\dfrac{N}{N_o}=e^{-0.0178\times 350.4}[/tex]
[tex]\dfrac{N}{N_o}=0.0019[/tex]
Hence, this is the required solution.
