Đáp án:

a) Đ

b) Đ

c) S

d) S

Giải thích các bước giải:

a) Ta có:

\({}_{27}^{60}Co \to {}_{28}^{60}Ni + {\beta ^ - }\)

b) Ta có: 

\(\begin{array}{l}
{H_0} = \lambda {N_0} = \dfrac{{\ln 2}}{T}.\dfrac{m}{M}.{N_A}\\
 \Rightarrow {H_0} = \dfrac{{\ln 2}}{{5,27.365.86400}}.\dfrac{{15}}{{60}}.6,{02.10^{23}}\\
 \Rightarrow {H_0} = 6,{28.10^{14}}Bq
\end{array}\)

c) Số hạt nhân \({}_{28}^{60}Ni\) tạo thành = số hạt nhân \({}_{27}^{60}Co\) bị phân rã.

Ta có: 

\(\begin{array}{l}
{N_{Ni}} = \Delta {N_{Co}} = {N_{{0_{Co}}}}\left( {1 - {2^{ - \frac{t}{T}}}} \right)\\
 \Rightarrow {N_{Ni}} = \dfrac{{{m_{Co}}}}{{{A_{Co}}}}.{N_A}\left( {1 - {2^{ - \frac{t}{T}}}} \right)\\
{m_{Ni}} = \dfrac{{{N_{Ni}}}}{{{N_A}}}.60 = \dfrac{{\dfrac{{{m_{Co}}}}{{{A_{Co}}}}.{N_A}\left( {1 - {2^{ - \frac{t}{T}}}} \right)}}{{{N_A}}}.60\\
 \Rightarrow {m_{Ni}} = {m_{Co}}\left( {1 - {2^{ - \frac{t}{T}}}} \right) = 15\left( {1 - {2^{ - \frac{{7,25}}{{5,27}}}}} \right) = 9,22g
\end{array}\)

d) Ta có:

\(\begin{array}{l}
\dfrac{{{m_{Co}}}}{{{m_{Ni}}}} = \dfrac{{\dfrac{{{N_{Co}}}}{{{N_A}}}.60}}{{\dfrac{{{N_{Ni}}}}{{{N_A}}}.60}} = \dfrac{{{N_{Co}}}}{{{N_{Ni}}}} = \dfrac{{{N_0}{{.2}^{ - \frac{t}{T}}}}}{{{N_0}\left( {1 - {2^{ - \frac{t}{T}}}} \right)}} = \dfrac{{{2^{ - \frac{t}{T}}}}}{{1 - {2^{ - \frac{t}{T}}}}}\\
 \Rightarrow \dfrac{{{m_{Co}}}}{{{m_{Ni}}}} = \dfrac{{{2^{ - \frac{{2,56}}{{5,27}}}}}}{{1 - {2^{ - \frac{{2,56}}{{5,27}}}}}} = 2,5
\end{array}\)