Preparation problems

44-10) How much U-235 would have to fission to produce an explosive yiled of 1.0 kt?

From the reading, we know that 1kt is 4.18 x 10¹² J of energy or 2.61 x 10²⁵ MeV.
We also know that each U-235 fission releases about 200 MeV.
(2.61 x 10²⁵ MeV) / (200 MeV/fission) = 1.3 x 10²³ fissions or 1.3 x 10²³ nucleii.
Each nucleus has a mass of (235 u) x (1.66054 x 10^-27 kg/u) = 3.9 x 10^-25 kg.
(1.3 x 10²³ nucleii) x (3.9 x 10^-25 kg/nucleus) = 0.05 kg

44-17) The temperature inside a typical reactor core is 600K. What is the thermal speed of a neutron at this temperature?

½mv² = 3/2 kT
v = (3kT/m)^½
= (3kT/m)^½
= (3 * 1.38x10^-23J/K * 600K / 1.67493x10^-27 kg)^½
= (1.483 x 10⁷)^½
= 3.8 km/s

44-19) A neutron collides ellastically and head-on with a staionary deuteron in a reacotr moderated by heavy water. How much of its kinetic energy is transferred to the deuteron?

Because it is an elastic collision, we know that both energy and momentum are conserved, so
m_nv_ni = m_nv_nf + m_dv_df and
½m_nv_ni² = ½m_nv_nf² + ½m_dv_df²
From Eqn 11-9, we get
v_nf = v_ni (m_n - m_d) / (m_n + m_d)
v_nf / v_ni = (m_n - m_d) / (m_n + m_d)

since the mass of the deuteron is about 3 times the mass of the neutron, v_nf / v_ni = -2m_n / 4m_n = -½

so the final speed of the neutron is ½ the initial (the - sign just means that its in the opposite direction), which means that it has lost ½ of its initial kinetic energy. Since the collision is elastic, no energy is lost, so
the neutron transfers ½ of its energy to the deuteron.

44-24) It is generally considered impossible for a slow-neutron reactor to blow up tlike a bomb, even in the event of a runaway raection. To help confirm this, compare the time for the reaction rate to double in (a) an out of control reactor with prompt-neotron gerenration time of 100 µsec and a multiplication factor k=1.01 and (b) in a bomb with t=10 ns and k=1.5.

First we need to find out how many generations it takes for the number of fissions to double. Keep in mind though that Eqn 44-2 tells us the total number of fissions (summed over all generations). What we are looking for is in which generation the number of fissions is twice the original number. For every fission we start with at t=0, we will get k fissions after the first generation. In the second generation we have k² and so on. So on the nth generation we get kⁿ fissions.
2 = kⁿ
ln(2) = n ln(k)
n = ln(2) / ln(k)
for the reactor:
n = 0.693/ln(1.01) = 69.66
So it takes 70 generations for the reaction rate to double. Since each generation takes 100 µsec, the total time is 7 msec
for the bomb:
n = 0.693 / ln(1.5) = 1.7
So it takes the bomb about 2 generations for the reaction rate to double, and since each generation take only 10 nsec, this means the total time is 20 nsec
In other words, the reactor take about 350,000 times as long to "explode".

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Solutions translated from T_EX by T_TH, version 1.57.