Question 1:
What is meant by:
(a) radioactive decay
(b) half-life
(c) nuclear energy
Answer:
(a) A radioactive decay is a random and spontaneous process by which an unstable nucleus will decay by emitting radioactive radiation to become a more stable nucleus.
(b) The half life, T½ is the time taken for a sample of radioactive nuclei to decay to half of its initial number.
(c) Nuclear energy is the energy produced by reactions in atomic nuclei.
What is meant by:
(a) radioactive decay
(b) half-life
(c) nuclear energy
Answer:
(a) A radioactive decay is a random and spontaneous process by which an unstable nucleus will decay by emitting radioactive radiation to become a more stable nucleus.
(b) The half life, T½ is the time taken for a sample of radioactive nuclei to decay to half of its initial number.
(c) Nuclear energy is the energy produced by reactions in atomic nuclei.
Question 2:
The following shows the equation for a radioactive decay:
(a) Identify X and Y in the decay equation.
(b) How many α and β-particles will be released when 22286Rn decays to 21082Pb?
Answer:
(a) X is the helium nucleus or α-particle, Y is γ.
(b)
$$ { }_{86}^{222} \mathrm{Rn} \rightarrow{ }_{82}^{210} \mathrm{~Pb}+x{ }_2^4 \mathrm{He}+y_{-1}^0 e $$
$$ \begin{aligned} 222 & =210+4 x+0 \\ 222-210 & =4 x \\ 12 & =4 x \\ x & =\frac{12}{4} \\ & =3 \alpha-\text { particles } \end{aligned} $$
$$ \begin{aligned} 86 & =82+2 x-y \\ 86 & =82+2(3)-y \\ y & =82+6-86 \\ y & =2 \beta-\text { particles } \end{aligned} $$
The following shows the equation for a radioactive decay:
(a) Identify X and Y in the decay equation.
(b) How many α and β-particles will be released when 22286Rn decays to 21082Pb?
Answer:
(a) X is the helium nucleus or α-particle, Y is γ.
(b)
$$ { }_{86}^{222} \mathrm{Rn} \rightarrow{ }_{82}^{210} \mathrm{~Pb}+x{ }_2^4 \mathrm{He}+y_{-1}^0 e $$
$$ \begin{aligned} 222 & =210+4 x+0 \\ 222-210 & =4 x \\ 12 & =4 x \\ x & =\frac{12}{4} \\ & =3 \alpha-\text { particles } \end{aligned} $$
$$ \begin{aligned} 86 & =82+2 x-y \\ 86 & =82+2(3)-y \\ y & =82+6-86 \\ y & =2 \beta-\text { particles } \end{aligned} $$
Thus, 3 α-particles and 2 β-particles are released.
Question 3:
(a) Astatine-218 has a half-life of 1.6 s. How long will it take for 99% of the nucleus in one sample of astatine-218 to disintegrate?
(b) Radium-226 has a half-life of 1 600 years. What percentage of the sample of radium-226 will be left after 8 000 years?
Answer:
(a)
$$ \begin{aligned} & 100 \% \xrightarrow{1} 50 \% \xrightarrow{2} 25 \% \xrightarrow{3} 12.5 \% \xrightarrow{4} 6.25 \% \\ & \xrightarrow{5} 3.125 \% \xrightarrow{6} 1.5625 \% \xrightarrow{7} 0.78125 \% \text { (disintegrate } \approx 99 \% \text { ) } \end{aligned} $$
$$ \begin{aligned} \text { Total time } & =7 \times 1.6 \mathrm{~s} \\ & =11.2 \mathrm{~s} \end{aligned} $$
(b)
$$ \begin{aligned} &\text { Number of half-life, } n=\frac{8000}{1600}=5\\ &N=\left(\frac{1}{2}\right)^n N_0 \end{aligned} $$
$$ \begin{aligned} &\text { Percentage of the sample of radium−226 remains }\\ &\begin{aligned} & =\left(\frac{1}{2}\right)^5 \times 100 \% \\ & =3.125 \% \end{aligned} \end{aligned} $$
$$ \text { So, after } 5 T_{\frac{1}{2}} \text {, only } 3.125 \% \text { of the sample remains. } $$
(a) Astatine-218 has a half-life of 1.6 s. How long will it take for 99% of the nucleus in one sample of astatine-218 to disintegrate?
(b) Radium-226 has a half-life of 1 600 years. What percentage of the sample of radium-226 will be left after 8 000 years?
Answer:
(a)
$$ \begin{aligned} & 100 \% \xrightarrow{1} 50 \% \xrightarrow{2} 25 \% \xrightarrow{3} 12.5 \% \xrightarrow{4} 6.25 \% \\ & \xrightarrow{5} 3.125 \% \xrightarrow{6} 1.5625 \% \xrightarrow{7} 0.78125 \% \text { (disintegrate } \approx 99 \% \text { ) } \end{aligned} $$
$$ \begin{aligned} \text { Total time } & =7 \times 1.6 \mathrm{~s} \\ & =11.2 \mathrm{~s} \end{aligned} $$
(b)
$$ \begin{aligned} &\text { Number of half-life, } n=\frac{8000}{1600}=5\\ &N=\left(\frac{1}{2}\right)^n N_0 \end{aligned} $$
$$ \begin{aligned} &\text { Percentage of the sample of radium−226 remains }\\ &\begin{aligned} & =\left(\frac{1}{2}\right)^5 \times 100 \% \\ & =3.125 \% \end{aligned} \end{aligned} $$
$$ \text { So, after } 5 T_{\frac{1}{2}} \text {, only } 3.125 \% \text { of the sample remains. } $$