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19. Plum Pudding Model. An alpha particle (two protons, two neutrons) h heads di

ID: 2305932 • Letter: 1

Question

19. Plum Pudding Model. An alpha particle (two protons, two neutrons) h heads directl model of the nucleus. That is, we assume the nuclear radius to be the size of an atom: 1 as 24.6MeV of kinetic energy and y for a stationary gold nucleus (79 protons, 118 neutrons). Here we will use the plum pudding ) ssng the alpha begins very far away, determine its kinetic energy when it reaches the outside of the nucleus (r-R 10-1°m). (b) Now we want to determine the kinetic energy when the alpha gets to the center of the nucleus. For this part your answers should be algebraic. Use only: Qaljpa 2e, Qgold-79e, k, R and r in your answers i. Use gauss's law to determine the electric field inside a uniformly charged sphere of net charge ii. Use this information to make a graph of the force on the alpha as a function of its positionr iii. Determine the work done on the alpha as it moves from r R to r-0. The easiest way is to 79e and radius R. within the sphere. Label the maximum value on the y-axis (force axis). graphically determine the area under the curve. Is this work positive or negative?

Explanation / Answer

19. alpha particel, 2p, 2N,

m= 4mo

where mo = mp = mn

KE = 24.6 MeV

nucleus,

M = (79 + 118)mo = 197mo

a. at r = 10^-10 m

initial total energy = KE = 24.6 MeV

final KE = ?

final PE = k(2e*79e)/r

e = 1.6*10^-19 C

hence

final PE = 0.002270144 MeV

hence from conservation of energy

final KE = 24.6 - 0.002270144 = 24.597729856 MeV

b. a. for a sphere of radius R

charge 79e

from gauss law

electric field is at radius r

E(r) = k*79e*r/R^3

b. within the sphere

force = 2e*E(r)

hence following is the graph between force and distance

c. work done as alpha particel moves from r = R to r = 0 is W

dW = Fdr = 2e*k*79e*rdr/R^3

W = 2e*k*79e*R^2/R^3

W = 2e*k*79e/R = 36322.304*10^-20 J = 2270.144 eV = 2.270144 keV

c. at center of nucleus, KE = 24.597729856 - 2.270144/1000 MeV = 24.595459712 MeV

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