Question

For the fission reaction ${}^{\mathbf{235}}\mathbf{U}\mathbf{+}\mathbf{n}\mathbf{\to }\mathbf{X}\mathbf{+}\mathbf{Y}\mathbf{+}\mathbf{2}\mathbf{n}$. Rank the following possibilities for X (or Y), most likely first: ${}^{\mathbf{152}}\mathbf{Nd}$,${}^{\mathbf{140}}\mathbf{I}$, ${}^{\mathbf{128}}\mathbf{In}$, ${}^{\mathbf{115}}\mathbf{Pd}$,${}^{\mathbf{105}}\mathbf{Mo}$ .

Short answer

The ranking for possibilities of nuclei fragments in a fission reaction is ${}^{140}\mathbf{I}$, ${}^{\text{105}}\text{Mo}$, ${}^{\text{152}}\text{Nd}$, ${}^{\text{123}}\text{In}$ and ${}^{\text{115}}\text{Pd}$.

Step-by-step solution

Variation of percentage yield of nuclei with the mass number of fragments

The percentage yield of fragments in nuclear fission for different mass numbers is given below in figure 43-1.

The percentage yield of fragments is higher for the atoms which have mass numbers nearer to 95 and 140. The percentage yield for fragments is lowest for mass numbers nearer to 120.

Identification of nuclei for most likely in the given nuclear fission reaction

The percentage yield for fragments of Iodine is highest, so the possibility of X being Iodine is highest in given nuclei.

The mass number of Molybdenum is 105, which is nearer to 100, and its percentage yield is the second highest from given nuclei, so the second most likely possibility for X is Molybdenum.

The Neodymium is suitable for the most likely possibility of X because its percentage yield is more than the remaining two nuclei.

The mass number of Indium is 128, which is closer to 120, so the percentage yield is a little more than the lowest yield. It's possible to be X is fourth among the given nuclei.

The percentage yield corresponding to mass number 115 is almost the lowest, so the possibility of X being Palladium is last among the given nuclei.

Therefore, the ranking for possibilities of nuclei fragments in a fission reaction is ${}^{140}\mathrm{I}$, ${}^{105}\mathrm{Mo}$, ${}^{152}\mathrm{Nd}$, ${}^{123}\mathrm{In}$and ${}^{115}\mathrm{Pd}$.