Periodic table contains 11 transuranic actinide elements

[anshul04]

Hi guys,

Stumbled across this MSR question in one of the MGMAT CATs. The question (and the associated data) itself looks okay; however, answer explanations of two questions seem conflicting to me.

On one hand, in the explanation of question 1 (shown below), it is mentioned that we can actually derive the atomic number and mass number of an element after one half-cycle even if the element decays through two modes (provided one of the decay mode is SF because " SF has no effect on the mass number of an element").

On the other hand, in the explanation of question 2 (also shown below), it is mentioned that if an element has more than one possible decay mode, it is impossible to know its atomic number after one cycle of decay. Additionally, decay mode SF, since it breaks the nucleus apart entirely, changes the atomic number unpredictably.

Could anyone please help me in understanding this? Which explanation is correct and which one is not?

Source1:
The periodic table of chemical elements contains 11 transuranic actinide elements, with atomic numbers ranging from 93 to 103. An element is defined by its atomic number, the number of protons in the atomic nucleus. Nuclei with the same atomic number but different numbers of neutrons are isotopes of the same element. The number of possible isotopes that have been attested in the laboratory is listed in the table shown for each element. The mass number of an isotope is the total number of protons and neutrons. All of the transuranic actinides are unstable. For any element, the most stable isotope is the one with the longest half-life, or time until half of the original amount present decays. The possible decay modes of the most stable isotope include Alpha particle emission, which drops the atomic number by 2 and the mass number by 4; Spontaneous Fission (SF), in which the nucleus breaks into several fragments; Beta particle emission, which increases the atomic number by 1 but leaves the mass number unchanged; and Electron Capture (EC), which decreases the atomic number by 1 but leaves the mass number unchanged.

Source2:
Atomic | Element | Mass # | Smallest Mass# | Melting | Density
Num | | of most | of an attested | point(in | (kg/m^3)
| stable | isotope |Kelvins) | of MSI
| isotope | | of MSI |
93 Neptunium 237 225 910 20,450
94 Plutonium 244 228 913 19,816
95 Americium 243 231 1,449 n/a
96 Curium 247 232 1,613 13,510
97 Berkelium 247 235 1,259 14,780
98 Californium 251 237 1,173 15,100
99 Einsteinium 252 240 1,133 13,500
100 Fermium 257 242 1,800 n/a
101 Mendelevium 258 245 1,100 n/a
102 Nobelium 259 249 1,100 n/a
103 Lawrencium 262 251 1,900 n/a

Source3:
Atomic Num | Half Life | Time Unit of | Decay Mode
| Half-Life |
93 2,140,000 years Alpha; SF
94 82,000,000 years Alpha; SF
95 7,370 years Alpha; SF
96 15,600,000 years Alpha
97 1,400 years Alpha
98 898 years Alpha; SF
99 1.29 years Alpha; Beta; EC
100 100.5 days Alpha; SF
101 51.5 days EC
102 58 minutes Alpha; SF; EC
103 3.6 hours SF; EC

Question 1:
The most stable isotope of californium can decay directly to the most stable isotope of curium. - Yes / No

Explanation:
Yes. First, check the possible decay modes of californium. They are listed as Alpha particle emission and Spontaneous Fission (SF). Now, we must turn back to the descriptions of the various decay modes provided in the first tab. SF has no effect on the mass number of an element, which leaves only Alpha particle emission. This decay mode is described as causing an element to drop "the atomic number by 2 and the mass number by 4." The atomic number of californium is 2 greater than the atomic number of curium, and the mass number is 4 greater. This fits the description of the decay mode exactly.

Question 2:
The atomic number of an atom that was once neptunium but that has decayed through one half-life cycle. - Can be determined / Can not be determined?

Explanation:
Cannot Be Determined. Neptunium (#93) has two decay modes. If an element has more than one possible decay mode, it is impossible to know its atomic number after one cycle of decay (because three of the four decay modes change the atomic number in three different ways, and the fourth, spontaneous fission or SF, breaks the nucleus apart entirely, changing the atomic number unpredictably). So, to know the result, we’d need to be sure that the element had only one decay mode (and that decay mode can’t be SF).

P.S. - Please bear with me for bad-formatting of the tables' data; I could not find an easy way to insert screenshots in the post. And although, I spent considerable time trying to make this data as easy to understand as possible, it still looks as bad as it was in the beginning (when I just copy-pasted it from the question).

[tim]

Thanks for bringing this up. I believe that in this part of the explanation to question 1:

"SF has no effect on the mass number of an element"

the "no effect" should be changed to "an unknown/unpredictable/indeterminate effect"

If that change were made, would that clear up the confusion you are having? To say that SF has "no" effect is certainly incorrect, because the information provided in the setup for the problem contradicts this.

[anshul04]

Thank you Tim for your response!

Well, that would definitely help; but in that case, answer of question 1 should be changed to No (from Yes).

If we agree on the fact that "SF has an unknown/unpredictable/indeterminate effect on the mass number of an element", then we can not determine the exact mass number & atomic number of Californium after its one half-cycle decay and hence, we can not say for sure that the most stable isotope of Californium would get decayed directly to Curium.

It would be on the same line as the explanation of answer of question 2 is.

Please let me know if my understanding is not correct.

[tim]

Question 1 says "can", and if the decay involves an alpha particle, then this situation CAN occur. Does that help?

[anshul04]

Although I am still not fully convinced, I guess I will let it go.

Thanks again Tim!

[RonPurewal]

Glad we could help.