how to calculate MeV

[james-p]

to calculate the energy released from a deutiriume tritium fusion to creat helium4 with a neutron do take the MeV from deutirium and add it to tritium the substact the MeV from He4 and the total left would be the energy your gaining?

ex H2 MeV is 1.2 + H3 MeV is 2.9 =4.1 and He4 MeV is 7.1 so 7.1 - 4.1 = 3

soo 3 is he amount of energy your liberating? is this right or is their a difrent formula?

and in this link how does he get difrent binding energy compared to the graphic?

thanks

james

[swansont]

In the video the binding energy values are 2 MeV, 8 MeV and 28 MeV, but the approach is right. The difference in the binding energy of the reactants and products tells you the energy released or required for the reaction to occur. You get the BE by reading off the graph or by taking the mass of the nucleons and subtracting the mass of the resulting nucleus, and multiplying it all by c^2

[james-p]

but wher does he get 2MeV 8MeV and 28MeV

thanks

[swansont]

but wher does he get 2MeV 8MeV and 28MeV

thanks

 

From the graph, or from the calculation I described.

[james-p]

but on the graph it shows that deutirium is 1.2 mev and not 2 and tritium 2.9 mev not 8

 

did i solv the equation right if not could you show me how to thnx

2.0136 + 3.0160492 = 4.002602 + 1.008647668

 

D T He-4

 

5.0296492 = 5.011249668

 

D+T He-4 + N

 

5.0296492 - 5.011249668 = 0.018399532

 

D+T He-4 + N

 

E=MC^2

 

=0.00000000000000000000000000003055kg x (3x10^8ms^-1)^2

 

D T c^2

 

=0.00000000000000000000000000003055kg x (3x 10^8ms^-1) x (3x10^8ms^-1)

 

=0.00000000000000000000000000003055kg x (9x10^16m^2s^-2)

 

=0.00000000000000000000000000003055kg x (9x10^16) kg m^2 s^-2

 

= 0.00000000000000000000000000003055kg x 90000000000000000 kg m^2 s^-2

 

=0.0000000000027495MeV

 

 

[swansont]

but on the graph it shows that deutirium is 1.2 mev and not 2 and tritium 2.9 mev not 8

 

did i solv the equation right if not could you show me how to thnx

2.0136 + 3.0160492 = 4.002602 + 1.008647668

 

 

The curve is binding energy per nucleon — you have to multiply by the number of nucleons, as the video instructed. The numbers were approximate, but deuterium has 2 nucleons, so the total BE is ~ 2 MeV.

 

For the calculation, those numbers are in atomic mass units. You can go directly into MeV by using c^2 = ~931.5 MeV/amu

[james-p]

ok so if i have a mass of 0.00000000000000000000000000003055kg how can i convert this into MeV energy using E=MC2

could you show me the calcules becaus idont think i did it the right way thanks

[swansont]

There are Avogadros number of atomic mass units in a gram, or 6.02 x 10^26 in a kg

[james-p]

soo how do you solve the equation when i put the number in it tells me their is only 0.000000000002745697071845 joules and not 18 MeV

 

[swansont]

soo how do you solve the equation when i put the number in it tells me their is only 0.000000000002745697071845 joules and not 18 MeV

 

You could take the definition of electron-Volt and do a unit conversion. (1 Volt = 1 Joule/Coulomb) Do you know what the fundamental charge is, in Coulombs?

[james-p]

thanks i finally figures it out is this right

E=MC^2

 

= 0.00000000000000000000000000003055kg x (299792458)^2m/s

 

=0.00000000000000000000000000003055kg x 89875517900000000 m/s

 

=0.000000000002745697071845 J

 

And 0.00000000000016021765 J equals to 1 MeV so 0.000000000002745697071845 J / 0.00000000000016021765 J equals to

 

17.137294622939482635028038421485 MeV.

[swansont]

Your next step should be to learn about significant digits in calculations.

[james-p]

alredy did,all digits after the last number are ussless

[swansont]

= 0.00000000000000000000000000003055kg x (299792458)^2m/s

 

You only haver four significant digits here, so you should only have four at the end.

 

(Also, learning scientific notation will ultimately help you in calculations. e.g. 3.055 x 10^-29 kg Dropping or adding a leading zero is really easy, and will give you the wrong answer.)

[james-p]

so it sould be

3.055 x 10^-29 kg x

2997 x 10^5 =

9.155^-21

 

 

[swansont]

Pretty much. c could be expressed as 2.997 x 10^8; you usually only have one number to the left of the decimal point, but it's not absolutely required. It's a huge help if you don't have a calculator handy, since the exponents merely add together.

[james-p]

ok thanks