Consider the simplified model of an atomic nucleus as a uniform ball of charge w

Question

Consider the simplified model of an atomic nucleus as a uniform ball of charge with volume propotional to the number of nucleons (protons and neutrons combined), and the volume required per nucleon is 0.9 fm^3 (regardless of whether it is a proton or neutron). (a) Calculate the electric field at the surface of the uranium-238 nucleus. (b) Would the electric field be larger or smaller for the uranium-235 nucleus? What about for the thorium-238 nucleus? (c) If there is a general trend in atomic nuclei that approximately half of the nucleons are protons and half neutrons, then prove that the electric field at the surface of the nucleus tends to increase as the size of the nucleus increases. The figure represents a dipole (two equal and opposite charges plusminus q separated by a distance d) along the y axis, centered at the origin. (a) Derive equations for the x and y components of the electric field at an arbitrary point on the x axis as a function of position. (b) Calculate the electric field (magnitude and direction) at (x, y) = (1, 0) m if P = qd = 1 mu C-mm, and d

Explanation / Answer

8. given, Volume of nucleus proportional to n where n is number of nucleons

so if radius of nucleus is r

4pi*r^3/3 = kn [ where k is a constant]

so volume required per nucleon = 0.9 *fm^3 = 0.9*10^-45 m^3

a. for U238

Volume of nucleus = 0.9*10^-45*238 = 214.2*10^-45 m^3

radius of nucleus = r

4pi*r^3/3 = 214.2*10^-45

r = 3.7117*10^-15 m

electric field on surface of nucleus = ke*A/r^2 [ A is atomic number of element and e is charge on proton]

E = 8.98*10^9*(1.6*10^-19)*92/r^2 = 95.948*10^20 V/m

b. at surface of U235

radius of nucles = r'

4*pi*r'^3/3 = 0.9*10^-45*235

r' =3.696*10^-15 m

E = ke*A/r^2 = 8.98*10^9*(1.6*10^-19)*92/r^2

as r is lesser but A is the same, so the field is stronger

for Th 238, r is same but A is moreso E is stronger too

c. if half of nucleons are protons

then for a radius r of nucleus

4*pi*r^3/3 = 0.9*10^-45*(2A)

r = 7.546*10^-16 A^1/3 [ wher A is atomic number]

spo electric field on surtfac eof nucleus = kA*e/r^2 = 25.2*10^20*A/A^2/3 = 25.2*10^20*A^1/3

so as A increases, size increases and so does E

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