The radius of a 231 Th (thorium) nucleus (90 protons and 141 neutrons) is nuc 75

Question

The radius of a 231 Th (thorium) nucleus (90 protons and 141 neutrons) is nuc 75x10 15 m. A helium nucleus (2 protons and 2 neutrons, m 6.6x10 231Th nucleus 27 kg) has been ejected from the nucleus in a transmutation process called alpha decay, with just enough speed that it escapes the attractive pull of the positive nucleus. What is the minimum kinetic energy of the helium nucleus? aw to find the electric field at the surface of the 23 Th nucleus Step 1: Use Gauss a. Sketch a spherical Gaussian surface of radius r around the nucleus, close to its surface. (This is already done for you in the figure.) b. Written as a multiple of e, how much charge is enclosed inside the surface? c. Assuming the charge to be uniformly distributed throughout the nucleus, what is the direction of the electric field relative to the surface normal of the Gaussian surface? its magnitude the same everywhere on the Gaussian surface? d. Write down Gauss' law (copy Eq. 27.18 here) e. What is the integral equal to? (You do not need to evaluate any integral. E is constant and comes out of the integra dA simply equals the area of the Gaussian surface.) f Substitute the results of Steps 1.b and 1.e into the equation that you wrote for Step 1.d. g. Solve for E(r)

Explanation / Answer

part b:
charge enclosed=90*charge on each proton=90*e

part c:

electric field will be radially outward as the charge on protons is positive.

part d:

gauss' law states that total electric flux passing through a surface is equal to charge enclosed.

if electric field is E, then integration of

epsilon*E*ds=charge enclosed

part e:

integral is equal to epsilon*E*4*pi*r^2

part f:

so equating charge enclosed to 90*e

epsilon*E*4*pi*r^2=90*e

part g:

E=90*e/(4*pi*epsilon*r^2)

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