I have an isotope,

292
uuh
116

this suffers seven successive alpha decay reactions, I know each for alpha decay reaction the isotope loses

4
He
2

so the atomic weight decreases by four, and the atomic number by 2...

None of the elements shown in the periodic table show an element with an atomic weight for the isotope here

which after 7 alpha reactions I have calculated to

264
No
102

is this right ???

all the stuff in the periodic tables shows a different number any advice please ..

Freddo

I answered this came question from someone about 3-4 days ago. The 102No^264 is correct. Only a few atoms of Uuh have been made and most of the chemistry is unknown. I expect that is why you don't see this particular isotope mentioned. But if you KNOW it goes through 7 alpha emissions, then this must be correct. How do you know it goes through 7 alpha emissions?

Hi

The question asks that you suppose this isotope undergoes a succession of seven alpha decay reactions.. and asks that you describe the final product which is an actinide...

Thanks

I hope you're correct! Otherwise, back to the drawing board for me!!!

To verify whether the final product after seven alpha decay reactions is indeed 102No^264 as you calculated, you can follow a few steps:

1. Start with the original isotope: 292uuh116.
2. Each alpha decay reaction involves the emission of a helium nucleus (4He^2+) and results in a decrease of 4 in atomic weight (mass number) and a decrease of 2 in atomic number (atomic symbol).
3. Perform seven successive alpha decay reactions, keeping track of the changes in atomic weight and atomic number after each decay.

Let's go through the process step by step:

The original isotope:
292uuh116

First alpha decay:
- Emission of an alpha particle (4He^2+)
- Atomic weight decreases by 4: 292 - 4 = 288
- Atomic number decreases by 2: 116 - 2 = 114
Resulting isotope: 288uut114

Second alpha decay:
- Emission of an alpha particle (4He^2+)
- Atomic weight decreases by 4: 288 - 4 = 284
- Atomic number decreases by 2: 114 - 2 = 112
Resulting isotope: 284uub112

Third alpha decay:
- Emission of an alpha particle (4He^2+)
- Atomic weight decreases by 4: 284 - 4 = 280
- Atomic number decreases by 2: 112 - 2 = 110
Resulting isotope: 280uus110

Fourth alpha decay:
- Emission of an alpha particle (4He^2+)
- Atomic weight decreases by 4: 280 - 4 = 276
- Atomic number decreases by 2: 110 - 2 = 108
Resulting isotope: 276uun108

Fifth alpha decay:
- Emission of an alpha particle (4He^2+)
- Atomic weight decreases by 4: 276 - 4 = 272
- Atomic number decreases by 2: 108 - 2 = 106
Resulting isotope: 272uub106

Sixth alpha decay:
- Emission of an alpha particle (4He^2+)
- Atomic weight decreases by 4: 272 - 4 = 268
- Atomic number decreases by 2: 106 - 2 = 104
Resulting isotope: 268uuj104

Seventh alpha decay:
- Emission of an alpha particle (4He^2+)
- Atomic weight decreases by 4: 268 - 4 = 264
- Atomic number decreases by 2: 104 - 2 = 102
Resulting isotope: 264No102

Therefore, based on your calculations, the final product after seven successive alpha decay reactions is indeed 264No^102. It is possible that this isotope is not explicitly mentioned in some periodic tables due to its relatively low occurrence or limited experimental data available for some extremely heavy and unstable elements. However, your calculations are consistent with the principles of alpha decay, so your answer seems to be correct.