Question

${ }_{92}^{238} U$ has $92$ protons and $238$ nucleons. It decays by emitting an alpha particle and becomes:

• A. ${ }_{92}^{234} U$
• B. ${ }_{90}^{234} Th$
• C. ${ }_{92}^{235} U$
• D. ${ }_{93}^{237} Np$

Answer 1

Answer: (B)

${ }_{90}^{234} Th$

Answer 2

As a general rule in any decay sum of mass number $A$ and atomic number $Z$ must be same on both sides. Let the daughter nucleus be ${ }_{ Z }^{ A } X$. So, reaction can be shown as ${ }_{92}^{238} U \longrightarrow{ }_{ Z }^{ A } X +{ }_{2}^{4} He$ From conservation of atomic mass $238 = A +4$ $\Rightarrow A =234$ From conservation of atomic number $92 = Z +2$ $\Rightarrow Z =904$ So, the resultant nucleus is ${ }_{90}^{234} X$, i.e., ${ }_{90}^{234} Th$. Note: A nuclide below the stability line in uranium decay series disintegrates in such a way that its proton number decreases and its neutron to proton ratio increases. In heavy nuclides this can occur by alpha emission.