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Topic: the additional proton  (Read 11405 times)

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Offline khwcm

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the additional proton
« on: June 02, 2010, 07:13:15 AM »
Observe that
4α + 9Be  14N + -1β
so,
for the "reactant"(alpha and Be)
there are  6 protons & 7 neutrons together

while for "products"(N and beta)
there are 7 protons and 7 neutrons

why there is a gain of one protons?

Offline AWK

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Re: the additional proton
« Reply #1 on: June 02, 2010, 08:32:55 AM »
4α + 9Be →12C + 1n
This reaction is used as small neutron source.

Your reaction is evidently wrong
AWK

Offline khwcm

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Re: the additional proton
« Reply #2 on: June 02, 2010, 09:51:34 AM »
thats just an example...
if u want real case, i have alots..
1427Si  :rarrow: 1327Al + e
1940K + e :rarrow: 1840Al

these are those cases with "initial proton No. and neutron No." different from the products one...
i have some more examples..

Offline DrCMS

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Re: the additional proton
« Reply #3 on: June 02, 2010, 10:20:44 AM »
thats just an example...
if u want real case, i have alots..
1427Si  :rarrow: 1327Al + e
1940K + e :rarrow: 1840Al

these are those cases with "initial proton No. and neutron No." different from the products one...
i have some more examples..


No these are quite different an unlike the first one they are correct

1427Si  :rarrow: 1327Al + e   this is beta decay  http://en.wikipedia.org/wiki/Beta_decay

1940K + e :rarrow: 1840Al   this is electron capture  http://en.wikipedia.org/wiki/Electron_capture

In your first post the total number of nucleons (protons+neutrons) on either side of the equation were different but in the last two examples the total number of nucleons on either side of the equation is the same.

Offline khwcm

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Re: the additional proton
« Reply #4 on: June 02, 2010, 12:25:40 PM »
but still, for the 2 equation:
just take the first one:
1427Si  :rarrow: 1327Al + e   
protons initial: 14, neutron = 13
protons final: 13, neutron = 14


Offline sjb

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Re: the additional proton
« Reply #5 on: June 02, 2010, 12:54:57 PM »
At a simple level, a neutron is decaying to give a proton and an electron. No contradiction here

Offline AWK

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Re: the additional proton
« Reply #6 on: June 04, 2010, 04:13:22 AM »
Quote
184018Al

Al ???
AWK

Offline DrCMS

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Re: the additional proton
« Reply #7 on: June 04, 2010, 07:43:22 AM »
 :P I'd missed that, should be Ar

Offline khwcm

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Re: the additional proton
« Reply #8 on: June 05, 2010, 03:36:30 AM »
At a simple level, a neutron is decaying to give a proton and an electron. No contradiction here

so is that a proton stored more energy than neutron?

Offline khwcm

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Re: the additional proton
« Reply #9 on: June 06, 2010, 04:25:15 AM »
At a simple level, a neutron is decaying to give a proton and an electron. No contradiction here

umum...one more question
for
1427Si   :rarrow: 1327Al + β
No. of electrons in Si is 14, and that in Al is 13

so, is that the β should come from Si ?
and the proton is just "converted" to a neutron, instead of giving out a neutron and electron?
btw, just interested to know, why the proton will convert to a neutron instead of forming a positively charged Al?

Offline vmelkon

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Re: the additional proton
« Reply #10 on: September 01, 2010, 01:05:53 PM »
"so is that a proton stored more energy than neutron?"

No, the entire nucleus has a lot of energy and by chance, one of the protons turns into a neutron and emits a positron and neutrino.
It is as if you think that beta radiation means an electron is emitted. There are in fact 2 types of beta : beta+ which is a positron and beta- which is an electron.

"27Si = 27Al + beta"

The orbiting electrons don't disappear so the Al atom would have a negative charge.

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