My Math Forum Nuclear reaction equation

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 May 12th, 2019, 12:05 PM #1 Senior Member   Joined: Mar 2019 From: TTF area Posts: 106 Thanks: 1 Nuclear reaction equation 23592U +, 1n0 →72 Zn30 +160 Sm X+ Y 1n0+energy Sorry for the formatting in advanced. Two questions What are the numbers represented by x and Y for X I'm thinking 62 for Y I'm thinking 4 Second question why is energy produced in this reaction?
May 12th, 2019, 01:02 PM   #2
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 Originally Posted by helpmeddddd 23592U +, 1n0 →72 Zn30 +160 Sm X+ Y 1n0+energy Sorry for the formatting in advanced. Two questions What are the numbers represented by x and Y for X I'm thinking 62 for Y I'm thinking 4 Second question why is energy produced in this reaction?
X represents the atomic number of Samarium (Sm) and Y represents the number of neutrons freed by the nuclear reaction.

$^{235}_{92}\text{U} + \, ^1_0\text{n} \to \, ^{72}_{30} \text{Zn} + \, ^{160}_{62} \text{Sm} + 4 ^1_0 \text{n} + \text{E}$

The energy is generated by the change in mass (mass defect) caused by the reaction ... $E = mc^2$

Further explanation ... https://www.thoughtco.com/definition...-defect-605328

May 12th, 2019, 01:19 PM   #3
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 Originally Posted by skeeter X represents the atomic number of Samarium (Sm) and Y represents the number of neutrons freed by the nuclear reaction. $^{235}_{92}\text{U} + \, ^1_0\text{n} \to \, ^{72}_{30} \text{Zn} + \, ^{160}_{62} \text{Sm} + 4 ^1_0 \text{n} + \text{E}$ The energy is generated by the change in mass (mass defect) caused by the reaction ... $E = mc^2$ Further explanation ... https://www.thoughtco.com/definition...-defect-605328
Thanks so In other words I've correctly identified the values of X and Y?

 May 12th, 2019, 01:33 PM #4 Senior Member   Joined: Mar 2019 From: TTF area Posts: 106 Thanks: 1 Calculate the mass deficit in this nuclear reaction 12C+\frac{12}{6}C\rightarrow \frac{24}{12}Mg Information given: rest mass of a carbon-12 nucleus: 1.9921157x10^{-26} rest mass of a magnesium nucleus: 3.9817469x10^{-26} rest mass of a proton: 1.67353x10^{-27} rest mass of a neutron: 1.67492x10^{-27} speed of light= 3.00x10^{8} ms^{-1}
May 13th, 2019, 05:35 AM   #5
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 Originally Posted by helpmeddddd Thanks so In other words I've correctly identified the values of X and Y?
The answer can be obtained by using the following principles:

1. Conservation of charge (in units of $\displaystyle e$)
2. Conservation of nucleon number

These are always conserved in nuclear reactions. There's other conservation laws too (like conservation of lepton number).

Mass and energy are also conserved together, but mass-energy conversions take place which can be a bit fiddly to work with (hence the E term).

So, let's take a look at the original equation:

$\displaystyle ^{235}_{92}\text{U} + \, ^1_0\text{n} \to \, ^{72}_{30} \text{Zn} + \, ^{160}_{X} \text{Sm} + Y ^1_0 \text{n} + \text{E}$

As Skeeter said, X can be identified by looking up samarium in the periodic table and getting its atomic number. However, for the sake of argument, let's pretend we don't know if it's samarium... we can still find X using the conservation laws:

1. Conservation of charge:

$\displaystyle 92 + 0 = 32 + X$

Therefore
$\displaystyle X = 92 - 32 = 60$

So even if the element's symbol were not written down, we could still figure out what element it is.

2. Conservation of nucleon number

$\displaystyle 235 + 1 = 72 + 160 + Y$

Therefore
$\displaystyle Y = 235 + 1 - 72 - 160 = 4$

It must be four neutrons.

As for calculating the energy, be careful about just using constituent rest masses because the actual measured mass of an atom is not equal to its constituents... that's because a there's also the binding energy of those atoms.

You need to basically solve this equation:

$\displaystyle m_{^{235}U} + m_n = m_{^{72}Zn} + m_{^{160}Sm} + 4m_n = E$

for E. You can either look up the masses of those atoms or look up the mass deficits and use them with the rest masses of their constituents to calculate their mass. Also make sure you use consistent units (It is normal to express all masses in energy units, such as MeV).

Last edited by Benit13; May 13th, 2019 at 05:37 AM. Reason: typos

May 13th, 2019, 11:47 AM   #6
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 Originally Posted by Benit13 The answer can be obtained by using the following principles: $\displaystyle 92 + 0 = 32 + X$ Therefore $\displaystyle X = 92 - 32 = 60$ So even if the element's symbol were not written down, we could still figure out what element it is. .
Where did you get 32 from? I did 92-30 to =62 how'd you get 60?

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