Question

Among the following, which has the longest half of life?
A. ${}_{90}^{232}Th$
B. ${}_{93}^{237}Np$
C. ${}_{92}^{238}U$
D. ${}_{92}^{235}U$

Answer 1

Hint: For an element to have a long half life, they need to be the most stable out of the given options. Unstable Nuclei will have an odd number of neutrons and odd number of protons. Stable Nuclei will have even number of neutron and even number of protons In case two elements have even number of neutron and proton then their $\dfrac{{{N_{neutron}}}}{{{N_{proton}}}} \sim 1$.

Complete step by step answer:
For any Nuclei to be stable, it should not undergo decay. If the Nuclei has an odd mass number it is less stable than the Nuclei which have an even mass number.

Odd mass numbers can happen for two reasons, either both the number of neutrons and protons is odd or any one of them is odd.

We will look at each option and compare their stability by their number of Neutrons and Protons.

Since we know, ${N_{Neutron}} = MassNumber - {N_{proton}}$
${N_{Neutron}}$= Number of neutron ${N_{proton}}$= Number of protons

$Mass\,number = 232,\,\,{N_{proton}} = 92$
${}_{90}^{232}Th$, Thorium has $90$ electrons, which means it also has $90$ protons, and hence the number of neutrons are $232 - 90 = 142$.
It has an even number of protons and Neutrons and hence, it is stable.

$Mass\,number = 237,\,\,{N_{proton}} = 93$
${}_{93}^{237}Np$,Neptunium has $93$electrons, which means it also has $93$protons, and hence the number of neutrons are $237 - 93 = 144$.
It has an even number of neutrons but an odd number of protons and hence it is less stable.

${}_{92}^{238}U$, Uranium has $92$ electron, which means it has $92$ protons, and hence the number of neutrons are $238 - 92 = 146$.
It has an even number of neutrons and protons and hence it is stable.

${}_{92}^{235}U$, Uranium has $92$ and hence it has $92$ protons, the number of neutrons are $235 - 92 = 143$. It has an odd number of neutrons and even number of protons and hence it is less stable.

Out of the Above option only, ${}_{92}^{232}Th$ and ${}_{92}^{238}U$ have both neutrons and protons odd. Out of these two, the most stable nuclei will have their $\dfrac{{{N_{neutron}}}}{{{N_{proton}}}} \sim 1$

For, ${}_{90}^{232}Th$ then\[\dfrac{{{N_{Neutron}}}}{{{N_{proton}}}} = \dfrac{{142}}{{90}} \Rightarrow 1.577\]

and for ${}_{92}^{238}U$ then $\dfrac{{{N_{neutron}}}}{{{N_{proton}}}} = \dfrac{{146}}{{92}} \Rightarrow 1.586$

Out of the above two, ${}_{90}^{232}Th$ has its $\dfrac{{{N_{neutron}}}}{{{N_{proton}}}} \sim 1$ as compared to ${}_{92}^{238}U$.
Hence, ${}_{90}^{232}Th$ is more stable.

So, Option A is correct.

Note: In case when either neutrons are odd or protons are odd and the other one is even, in such a case their stability will be determined by using the $\dfrac{{{N_{neutron}}}}{{{N_{proton}}}}$ ratio. The maximum stability will be for elements with both the neutron and proton number as even.