A) \[_{92}^{234}U\]
B) \[_{90}^{234}Th\]
C) \[_{92}^{235}U\]
D) \[_{93}^{237}Np\]
Correct Answer: B
Solution :
Key Idea: 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 \[_{92}^{238}U\xrightarrow{{}}_{Z}^{A}X+_{2}^{4}He\] From conservation of atomic mass \[238=A+4\] \[\Rightarrow \] \[A=234\] From conservation of atomic number \[92=Z+2\] \[\Rightarrow \] \[Z=90\] 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 nudides this can occur by alpha emission.
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