Đáp án:

a) Đ

b) S

c) S

d) Đ

Giải thích các bước giải:

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

b) Ta có: 

\(\begin{array}{l}
{N_0} = \dfrac{{{m_0}}}{A}.{N_A} = \dfrac{{15}}{{60}}.6,{023.10^{23}} = 1,{51.10^{23}}\\
\lambda  = \dfrac{{\ln 2}}{T} = \dfrac{{\ln 2}}{{5,27.365.24.3600}} = 4,{17.10^{ - 9}}\\
{H_0} = \lambda {N_0} = 4,{17.10^{ - 9}}.1,{51.10^{23}} = 6,{3.10^{14}}Bq
\end{array}\)

c) Ta có: 

\(\begin{array}{l}
{N_1} = {N_0}\left( {1 - {2^{ - \frac{t}{T}}}} \right) = 1,{51.10^{23}}\left( {1 - {2^{ - \frac{{7,25}}{{5,27}}}}} \right) = 9,{28.10^{22}}\\
{m_{Ni}} = \dfrac{{{N_0}}}{{{N_A}}}.A = \dfrac{{9,{{28.10}^{22}}}}{{6,{{023.10}^{23}}}}.60 = 9,24g
\end{array}\)

d) Ta có: 

\(\dfrac{{{N_1}}}{{{N_2}}} = \dfrac{{1 - {2^{ - \frac{t}{T}}}}}{{{2^{ - \frac{t}{T}}}}} = \dfrac{{1 - {2^{ - \frac{{2,56}}{{5,27}}}}}}{{{2^{ - \frac{{2,56}}{{5,27}}}}}} = 0,4\)