need help in quanta quarks question!! PLS HELP (1 Viewer)

erfanau123

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fusion of hydrogen atoms in the sun to form a helium nucleus can be represented by the equation

calculate the mass defect of the event and the energy produced


(specimen 4 marks )
 

chloe555

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so what don't you get about it? add up the individual masses of the protons, neutrons, and electrons on the left, then do the same for the right hand side, convert everything to amu. find the difference (this will be the mass defect) then use E=mc^2 to find the energy. all good?
 

packwolf

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They would have given you the actual masses of each.
 

erfanau123

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i tried and failed
i gave to 2 james ruse kids and they couldnt get it either

u try it dude!
 

b3kh1t

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Are the answers by any chance 1.987 u mass defect and 1.7883x10^17 J??

oops forgot to use amu sorry

sorry wrong equation is the energy released 1850.89 MeV
 
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HyperComplexxx

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1 amu = 1.661 × 10-27 kg
Mass of He = 4.003 amu = 6.647x10-27kg
Mass Proton = 1.673x10-27kg
Mass Neutron = 1.675x10-27kg

Mdefect = 2Mass Proton + 2Mass Neutron - Mass of He (there are 2 proton and 2 neutron in He)
= 4.9x10-29 kg
E = mchammer
= (4.9x10-29)(3x10^8)^2
= 4.41x10-12J ?

EDIT: Some post solution please :(
I am unsure of method to do it
EDIT2: Fixed it up a bit but probs wrong anyways
 
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nichorowitz

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yo cbf working it out but because it is actually atoms
just use the amu mass off the periodic table. find the difference then multiple by 931.5
this binding energy in MeV
EDIT:
actually didnt realise it was deuterium. that makes it a lot harder
 

erfanau123

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1 amu = 1.661 × 10-27 kg
Mass of He = 4.003 amu = 6.647x10-27kg
Mass Proton = 1.673x10-27kg
Mass Neutron = 1.675x10-27kg

Mdefect = 2Mass Proton + 2Mass Neutron - Mass of He (there are 2 proton and 2 neutron in He)
= 4.9x10-29 kg
E = mchammer
= (4.9x10-29)(3x10^8)^2
= 4.41x10-12J ?

EDIT: Some post solution please :(


wrong apparently, answers 3.8 x10^-12
I am unsure of method to do it
EDIT2: Fixed it up a bit but probs wrong anyways



Lol dude, umm yeh i got what u got but 3.8x10^-12 is correct, apparently the questions fukd cause they usually give u all info
 

b3kh1t

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Sorry, is this the right answer; mass defect 0.0024 u and the energy released 2.24 MeV

damn, solutions lol
 
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b3kh1t

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The question is stuffed, because the have to at least give you the mass of the deutrium, just realised, no need to worry Hypercomplexxx
 

weirdguy99

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The question is stuffed, because the have to at least give you the mass of the deutrium, just realised, no need to worry Hypercomplexxx
You find the mass of the deuterium by adding the constituent nucleons (2 protons + 2 neutrons for both nucleons).

Mass of helium is 6.646x10^-27
Mass of hydrogen is 6.688x10^-27
Defect is 4.2x10^-29

E=Defect x Light Squared
=3.8x10^-12J = 24 MeV
 
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b3kh1t

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You find the mass of the deuterium by adding the constituent nucleons (2 protons + 2 neutrons for both nucleons).

Mass of helium is 6.646x10^-27
Mass of hydrogen is 6.688x10^-27
Defect is 4.2x10^-29

E=Defect x Light Squared
=3.8x10^-12J = 24 MeV
how did you get that mass, I got 6.699x10^-27 (4sig. fig.) by adding the mass of the extra neutron??
 

FCB

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He got it out of the successone book.
 

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