1000 Solved Problems in Modern Physics

(Romina) #1

8.3 Solutions 447


8.92 Assuming that the elastic scattering of low energy neutrons is isotropic,
show that the mean energy of the neutron after each collision will be
Ef=(A^2 +1)Ei/(A+1)^2
whereAis the mass number of the target nucleus. Determine the number of
collisions needed to thermalize fission neutrons (2 MeV) in graphite (A=12).


8.2.15 Fusion ...........................................


8.93 Determine the range of neutrino energies in the solar fusion reaction,p+p→
d+e++ν. Assume the initial protons have negligible kinetic energy and that
the binding energy of the deuteron is 2.22 MeV,mp = 938 .3MeV/c^2 and
md= 1875 .7MeV/c^2 andme= 0 .51 MeV/c^2.


8.94 If the kinetic energy of the deuterons in the fusion reactionD+D→^32 He+
n+ 3 .2 MeV can be neglected, what is the kinetic energy of the neutron?
8.95 (a) It is estimated that the deutrons have to come within 100 fm of each other
for fusion to proceed. Calculate the energy that the deuterons must possess
to overcome the electrostatic repulsion.
(b) If the energy is supplied by the thermal energy of the deutrons, what is the
temperature of the deuteron?
[e^2 / 4 πε 0 = 1 .44 MeV fm, Boltzmann constantk= 1. 38 × 10 −^23 JK−^1 ]
(c) In (b) the actual required temperature is lower than the estimated value.
Explain the mechanism by which the fusion reaction may proceed.


8.96 In a fusion reactor, the D-T reaction withQvalue of 17.62 MeV is employed.
Assuming that the deuteron density is 7× 1018 m−^3 and the experimental value
<σDT·v>= 10 −^22 m^3 s
− 1
and that equal number of deuterons and tritons
exist in the plasma at energy 10 keV, calculate the confinement time if the
Lawson criterion is just satisfied.


8.3 Solutions


8.3.1 AtomicMassesandRadii ...........................


8.1 An ion of chargeqwill pick up kinetic energy,T=qVin dropping through a
P.D of V volts. In a magnetic inductionBperpendicular to its path, the ion of
momentumpwill describe a circular path of radius r given by
p=qBr=



2 MT=


2 MqV

M=

qB^2 r^2
2 V

(1)

For the first ion,q= 1. 6 × 10 −^19 C, B= 0. 08 ,r= 0 .0883 m and V=
400 V. Substituting these values in (1) we get
M 1 = 9. 98 × 10 −^27 kg
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