The stable end product in the uranium series \(\left({ }_{92}^{238} \mathrm{U}\right)\) is (a) \({ }_{82}^{205} \mathrm{~Pb}\) (b) \({ }_{82}^{206} \mathrm{~Pb}\) (c) \({ }_{82}^{207} \mathrm{~Pb}\) (d) \({ }_{82}^{208} \mathrm{~Pb}\)

Short Answer

Expert verified
The stable end product in the uranium series \({ }_{92}^{238} \mathrm{U}\) is \({ }_{82}^{206} \mathrm{~Pb}\). So the correct option is (b).

Step by step solution

01

Understanding the Uranium Decay Series

The Uranium 238 decay series refereed as the 'Uranium Series', is a chain of radioactive decays through which the uranium isotope will decay into a series of other isotopes before finally ending up as a stable isotope of lead.
02

Identifying the final Stable Isotope

In the Uranium Series, the stable end product is Lead-206. During the decay process, the atomic mass number decreases by 32 units (from 238 to 206) while the atomic number decreases by 10 units (from 92 to 82). This is accomplished through a series of alpha and beta minus decays.

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