Question

If 1 mg gold undergoes decay with 2.7 days half-life period, then find the amount left after 8.1 days is
A. 0.91 mg
B. 0.25 mg
C. 0.5 mg
D. 0.125 mg

Answer 1

Hint: According to law of radioactive disintegration if \[{N_0}\] is the initial number of atoms in a radioactive element and after time t remaining number of atoms is N for the element then, \[N = {N_0}{e^{ - \lambda t}}\] and \[\dfrac{N}{N_o}=\left(\dfrac{1}{2}\right)^n\] where, \[n=\dfrac{t}{t_{1/2}}\].

Formula used : \[\dfrac{N}{{{N_0}}} = {\left( {\dfrac{1}{2}} \right)^n}\] and \[n=\dfrac{t}{t_{1/2}}\]
Where, N = Number of atoms after time t, \[{N_0} = \] Initial number of atoms and \[{t_{1/2}} = \]Half life.

Complete step by step solution:
Given there is 1mg of gold that undergoes decay and has a half life \[{t_{1/2}}\]of 2.7 days, we have to find the amount of gold left after 8.1 days. According to law of radioactive disintegration, ratio between remaining amount (N) of radioactive element after time t and initial amount \[({N_0})\]of radioactive element is given by,
\[\dfrac{N}{{{N_0}}} = {\left( {\dfrac{1}{2}} \right)^n}\] Where, \[n=\dfrac{t}{t_{1/2}}\]
We have, N =?, \[{N_0} = \,1\,{\rm{mg}}\], t = 8.1 days, \[{t_{1/2}} = \,2.7\,{\rm{days}}\]
Therefore, \[n = \dfrac{{8.1}}{{2.7}} = 3\]
Then, \[\dfrac{N}{{{N_0}}} = {\left( {\dfrac{1}{2}} \right)^3}\]
As, \[{N_0} = 1\], therefore above equation will be,
\[ \Rightarrow N = \dfrac{1}{8}mg\, = 0.125\,mg\]
Hence, 0.125 mg of gold is left after 8.1 days.

Therefore, option D is the correct option.

Note: Isotope \[{}^{198}Au\] of gold has a half life of 2.697 days and undergoes beta-decay, because of its decay properties \[{}^{198}Au\] has been potentially used in nuclear medicine, radioactive tracing and nuclear weapons.