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bot dead
answer: C & D only
i dont get why A is false if
R = Ro * A^(1/3)
i'm not exactly sure what B is asking
and i don't seem to remember anything about D and E in NCERTWhy would A be true can you elaborate?
surface area proportional to R^2, which is proportional to A^2/3
wait but thats for the whole atom
That's for surface tension and drop thingy no?
surface tension ah?
nono
thats for the whole atom
the question specifies per nucleon
confused
okay hang on just try to do this from scratch
my thought process seems muddied
@Moderator hello guiz can one of u ping the Gyro Gearloose role
bot died when i posted this so the ping msg didnt come
@Gyro Gearloose
koi help kar dijiye aapka bhala ho jayega
Okay, as far as I have figured out:
A. Surface energy is proportional to area proportional to A^2/3. So surface energy per nucleon is proportional to A^(-1/3) ('Cuz of A nucleons)
B. Ig you have to assume uniform distribution of protons. The given option implies that all protons are in periphery of nuclei which is false. (Also dimensionally wrong as A^(1/3) hona chahiye tha ig)
D because of this
E maybe trivia, but I have no idea.
ahh
yeah B makes sense now
"coulomb contribution to the binding energy" = fancy phrase for electrostatic potential energy
however
i thought R = R_0 * A^1/3 was supposed to apply for the atom as a whole
hence the A^(1/3) correct term
That's for nuclei not an atom
nuclei sorry
i thought it was supposed to apply for the whole nucleus
not distances between individual nucleons
which are supposed to be whats contributing to potential energy
but yeah you can mark B off as dimensionally wrong
Well, B is dimensionally wrong. But yeah, I made an assumption that protons are uniformly distributed over the nuclei and visualize the nucleus as a drop of small +ve charges
yeah i came across some liquid drop model online while tryna figure this out
i think i'll do some further curiosity research on this part, but for now i'll mark as solved
thank u 👍
+solved @SirLancelotDuLac
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