
An artificial radioactive decay series begins with unstable ${}_{94}P{u^{241}}$. The stable nuclide obtained after eight $\alpha $-decays and five $\beta $-decays is
A. ${}_{83}B{i^{209}}$
B. ${}_{82}P{b^{209}}$
C. ${}_{82}T{i^{205}}$
D. ${}_{82}H{g^{201}}$
Answer
163.2k+ views
Hint:This problem is from the radioactivity section of physics. There are three types of radioactive decay. By applying the chemical equations of the alpha, beta and gamma decay, we can find the resultant stable nuclide.
Complete step by step solution:
The changes in atomic number and mass number after eight $\alpha $-decays.
$A = 241 - (8 \times 4) = 209$
$\Rightarrow Z = 94 - (8 \times 2) = 78$
The changes in atomic number and mass number after five $\beta $-decays.
$A = 209$ (No change in mass number)
\[\Rightarrow Z = 78 + (5 \times 1) = 83\]
The resultant stable nuclide will be \[{}_{83}{X^{209}} \Rightarrow {}_{83}B{i^{209}}\].
Hence, the correct option is option A.
Additional Information: The nucleus of an atom exhibits radioactivity due to nuclear instability. The decay rate of the nucleus is independent of temperature and pressure. Curie and Rutherford are the units of radioactivity. Radioactivity is dependent on the law of conservation of charge. The daughter nucleus has different physical and chemical properties than the parent nucleus.
The energy emitted by radioactivity is always accompanied by alpha, beta, and gamma particles.
1) Alpha decay: The decay of atomic nuclei with the emission of helium nuclei is called alpha decay. The alpha particle has two protons and two neutrons.
2) Beta decay: The decay of atomic nuclei with an electron or positron emission is called beta decay. A beta particle is generally referred to as an electron but it can also be a positron also.
3) Gamma decay: The decay of atomic nuclei with the emission of photons is called gamma decay. During the gamma decay process, neither A nor Z changes.
Note: Gamma rays are employed in radiotherapy because they kill cancerous cells. They are used to scan the inside of the body. Gamma rays kill bacteria in food and extend their shelf life by preventing decomposition.
Complete step by step solution:
The changes in atomic number and mass number after eight $\alpha $-decays.
$A = 241 - (8 \times 4) = 209$
$\Rightarrow Z = 94 - (8 \times 2) = 78$
The changes in atomic number and mass number after five $\beta $-decays.
$A = 209$ (No change in mass number)
\[\Rightarrow Z = 78 + (5 \times 1) = 83\]
The resultant stable nuclide will be \[{}_{83}{X^{209}} \Rightarrow {}_{83}B{i^{209}}\].
Hence, the correct option is option A.
Additional Information: The nucleus of an atom exhibits radioactivity due to nuclear instability. The decay rate of the nucleus is independent of temperature and pressure. Curie and Rutherford are the units of radioactivity. Radioactivity is dependent on the law of conservation of charge. The daughter nucleus has different physical and chemical properties than the parent nucleus.
The energy emitted by radioactivity is always accompanied by alpha, beta, and gamma particles.
1) Alpha decay: The decay of atomic nuclei with the emission of helium nuclei is called alpha decay. The alpha particle has two protons and two neutrons.
2) Beta decay: The decay of atomic nuclei with an electron or positron emission is called beta decay. A beta particle is generally referred to as an electron but it can also be a positron also.
3) Gamma decay: The decay of atomic nuclei with the emission of photons is called gamma decay. During the gamma decay process, neither A nor Z changes.
Note: Gamma rays are employed in radiotherapy because they kill cancerous cells. They are used to scan the inside of the body. Gamma rays kill bacteria in food and extend their shelf life by preventing decomposition.
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