# v. easy nuclear physics question

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#1
When radioactive isotopes decay, they sometimes have to go through a succession of disintegrations to reach a stable isotope. These are called decay chains, and involve the successive emission of numerous α and/or β particles.
One such isotope is radon-219 (219 Rn 86), which goes through a chain in which three α particles and two β particles are emitted before reaching a stable isotope.
What are the atomic and mass numbers of the resulting stable isotope?

My working: 219-12=207 (12 because 3 (3 a-particles) times 4 (mass). The answer says 215, though. Wrong mark scheme?
0
7 years ago
#2
(Original post by originaltitle)
When radioactive isotopes decay, they sometimes have to go through a succession of disintegrations to reach a stable isotope. These are called decay chains, and involve the successive emission of numerous α and/or β particles.
One such isotope is radon-219 (219 Rn 86), which goes through a chain in which three α particles and two β particles are emitted before reaching a stable isotope.
What are the atomic and mass numbers of the resulting stable isotope?

My working: 219-12=207 (12 because 3 (3 a-particles) times 4 (mass). The answer says 215, though. Wrong mark scheme?
One alpha particle is He (4, 2). So 3 would mean 12 subtracted from the mass number, which would give you 219-12=207. (don't worry why I'm writing this, I wanna work it out).
The atomic no would decrease by 6 and then increase by 2 due to β emission, so it would be 86-6+2=82.

I guess the mark sheet is wrong then. Do you have a snapshot of the original question?
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#3
(Original post by Dynamo123)
One alpha particle is He (4, 2). So 3 would mean 12 subtracted from the mass number, which would give you 219-12=207. (don't worry why I'm writing this, I wanna work it out).
The atomic no would decrease by 6 and then increase by 2 due to β emission, so it would be 86-6+2=82.

I guess the mark sheet is wrong then. Do you have a snapshot of the original question?
I do, actually. Question number 11: http://www.admissionstestingservice....-section-2.pdf.
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7 years ago
#4
The mark scheme says that c is the correct answer for 11, which is 207,82. So idk, maybe you misread it?
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#5
(Original post by InsertWittyName)
The mark scheme says that c is the correct answer for 11, which is 207,82. So idk, maybe you misread it?
My apologies.
0
7 years ago
#6
(Original post by Dynamo123)
One alpha particle is He (4, 2). So 3 would mean 12 subtracted from the mass number, which would give you 219-12=207. (don't worry why I'm writing this, I wanna work it out).
The atomic no would decrease by 6 and then increase by 2 due to β emission, so it would be 86-6+2=82.

I guess the mark sheet is wrong then. Do you have a snapshot of the original question?
Hi, thanks for posting the quesiton, it was very useful.
However I do not understand why the beta decay must always add one proton, I mean: beta decay could emit one positron (e+, I think) as well, therefore increasing the number of neutron and decreasing the figure for the protons.. or we should consider only the beta decay which emit electrons?
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7 years ago
#7
(Original post by LFFFF)
Hi, thanks for posting the quesiton, it was very useful.
However I do not understand why the beta decay must always add one proton, I mean: beta decay could emit one positron (e+, I think) as well, therefore increasing the number of neutron and decreasing the figure for the protons.. or we should consider only the beta decay which emit electrons?
Generally, we consider the Beta negative decay (i.e. emission of an electron), when we write "such a parent nuclide underwent a beta decya etc etc."
If a question concerning beta positive decay (with positron emission) comes up, it will be mentioned that the decay is to be taken as positron-emission.
0
7 years ago
#8
(Original post by Dynamo123)
Generally, we consider the Beta negative decay (i.e. emission of an electron), when we write "such a parent nuclide underwent a beta decya etc etc."
If a question concerning beta positive decay (with positron emission) comes up, it will be mentioned that the decay is to be taken as positron-emission.
It does, perfectly, thank you!
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7 years ago
#9
(Original post by LFFFF)
It does, perfectly, thank you!
Ah, no problem. ^.^
0
5 years ago
#10
I dont understand how loosing 2 elections (beta particles) causes the atmoic number to increase .. I though the atomic number was the number of protons, and so the number of elections, and loosing 2 elections causes -2 on the atomic number?
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