19. Plum Pudding Model. An alpha particle (two protons, two neutrons) h heads di
ID: 2305932 • Letter: 1
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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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