Physics 43!! How was it?? Expected GT anyone?

[Mattman]

will they cut marks for that?^because i did the same thing

 

[ninesone]

all you had to do is plug it into e=mc2, it was like 1.808u

hmm how about showing the binding energy per nucleon?
when i revised i thought i could easily solve these type of questions.. then stuck. :/

 

[jasters]

i dont know, i just did the question and got what they wanted

 

[jasters]

its just the mass defect that u calculate for, and the last question should have been 180mev

 

[jasters]

well something round that from what i can remember, hahaha yeah that sometimes happens to me also

 

[jasters]

will they cut marks for that?^because i did the same thing

maybe, i never worked back, i just did it the normal way

 

[Henry930821]

The hardest question was the one with the two graphs. Guys how did u all sketch for both the Hall voltage graph and the metal coil graph??

 

[Kandinsky]

its just the mass defect that u calculate for, and the last question should have been 180mev

was it for the energy released in the reaction?
I got 200 MeV (rhs mass - lhs mass)*c^2

 

[jasters]

they gave you the two binding energies in the table, and u cal the binding energy per nucleon for uranium , thus find the product of binding enrgy and number of total nucleons for each species, thee difference in Mev between the two sides of the equation is the energy released by the reaction, you cant just do the difference in mass between the two sides of the equation,, you had to use your answers in the previous questions for the final answer, so e=mc2 should not have been used in the final answer because the energies were already worked out in the previous questions, and i got 180 Mev and the other top boys in my school in physics got that, i think the main point in this topic is the mass defect of an atom, not the mass of an atom

 

[Kandinsky]

they also provided the masses of the products. and moreover, the energy was released in the reaction. while using your method, the energy on the RHS was bigger than that of the LHS. I think it's mass-energy which is conserved..

 

[jasters]

they also provided the masses of the products. and moreover, the energy was released in the reaction. while using your method, the energy on the RHS was bigger than that of the LHS. I think it's mass-energy which is conserved..

wait so are u saying i got it wrong or right?

 

[Kandinsky]

I don't know really. Maybe someone else will tell how to solve =)

 

[jasters]

wait so are u saying i got it wrong or right?

and did u get 7.18mev and 1.808u

 

[Kandinsky]

Yes

 

[jasters]

I don't know really. Maybe someone else will tell how to solve =)

well like the guy that got 95 in as got 180mev, and so did nost of the other people, but i am pretty sure of this because we had done heaps of these questions

 

[jasters]

i just dont get why u used mass because the energy's were already given, so you did not even have to use e=mc2

 

[Kandinsky]

http://www.thebigger.com/chemistry/...lculate-the-energy-released-nuclear-reaction/

take a look please. the masses were given not for nothing, i think

 

[jasters]

http://www.thebigger.com/chemistry/...lculate-the-energy-released-nuclear-reaction/

take a look please. the masses were given not for nothing, i think

they only gave the mass for uranium not the other two, they gave the binding energy PER NUCLEON for the other two, u get the product of 235 and 7.18MEV and do the same for the other two on the other side of the equation, the energy released is the diff between the two sides of the equation, what did other people get?

 

[Kandinsky]

they gave everything, including exact masses, maybe you didn't pay much attention.

 

[jasters]

they gave everything, including exact masses, maybe you didn't pay much attention.

and are u sure u did not just round up to 200