Question 462: Atoms & Nuclei

Question

In the uranium radioactive series, the initial nucleus is ${ }_{92}^{238} \mathrm{U}$ and that the final nucleus is ${ }_{82}^{206} \mathrm{~Pb}$. When uranium nucleus decays into lead, the number of $\alpha$-particles and $\beta$-particles emitted are

Accepted Answer

To determine the number of $\alpha$-particles and $\beta$-particles emitted when uranium-238 (${}_{92}^{238}\mathrm{U}$) decays into lead-206 (${}_{82}^{206}\mathrm{Pb}$), we must first understand the changes in atomic and mass numbers during the decay process.
Calculation of $\alpha$-particles
An $\alpha$-particle is composed of 2 protons and 2 neutrons, equivalent to a helium nucleus $({}_{2}^{4}\mathrm{He})$. Emitting one $\alpha$-particle decreases the atomic number by 2 and the mass number by 4.
Initial nucleus: ${}_{92}^{238}\mathrm{U}$
Final nucleus: ${}_{82}^{206}\mathrm{Pb}$
Change in mass number:
$$ 238 - 206 = 32 $$
Since each $\alpha$-particle reduces the mass number by 4:
$$ 	ext{Number of } \alpha	ext{-particles} = \frac{32}{4} = 8 $$
Calculation of $\beta$-particles
A $\beta$-particle decay involves the conversion of a neutron into a proton, increasing the atomic number by 1, but leaving the mass number unchanged.
Change in atomic number:
$$ 92 - 82 = 10 $$
To account for the 10-unit change in atomic number caused by the emission of 8 $\alpha$-particles (which decrease the atomic number by 16), $\beta$-particles must be emitted to increase the atomic number by 6 units (since $16 - 10 = 6$).
Hence, the number of $\beta$-particles is 6.
Conclusion
The decay of uranium-238 to lead-206 involves the emission of:
8 $\alpha$-particles
6 $\beta$-particles
Therefore, the correct answer is
Option D: $8 \alpha, 6 \beta$
.

Answer

$4 \alpha, 5 \beta$

Answer

$5 \alpha, 3 \beta$

Answer

$6 \alpha, 7 \beta$

Answer

$8 \alpha, 6 \beta$

Atoms & Nuclei

In the uranium radioactive series, the initial nucleus is { }_{92}^{238} \mathrm{U} and that the final nucleus is { }_{82}^{206} \mathrm{~Pb}. When uranium nucleus decays into lead, the number of \alpha-particles and \beta-particles emitted are

Choose the correct answer: