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Chapter 43: Q11Q (page 1330)

A nuclear reactor is operating at a certain power level, with its multiplication factor k adjusted to unity. If the control rods are used to reduce the power output of the reactor to 25% of its former value, is the multiplication factor now a little less than unity, substantially less than unity, or still equal to unity?

Short Answer

Expert verified

The multiplication factor for power output to 25% of its former value remains unity.

Step by step solution

01

Multiplication Factor

The ratio of the number of neutrons for a particular generation of power to the number of neutrons initially for the same power generation is called the multiplication factor.

02

Whether the multiplication factor is less than unity or equal to the unity

The multiplication factor for power output to 25% of its former value remains unity because the control rod provides a reduced number of neutrons. The number of neutrons provided by the control rod completes the deficiency of neutrons due to a reduction in power output.

Therefore, the multiplication factor for power output to 25% of its former value remains unity.

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Most popular questions from this chapter

(a) Calculate the rate at which the Sun generates neutrinos. Assume that energy production is entirely by the proton-proton fusion cycle. (b) At what rate do solar neutrinos reach Earth?

In an atomic bomb, energy release is due to the uncontrolled fission of plutonium Pu239(or U235). The bomb’s rating is the magnitude of the released energy, specified in terms of the mass of TNT required to produce the same energy release. One megaton of TNT releases 2.6×1028MeVof energy. (a) Calculate the rating, in tons of TNT, of an atomic bomb containing 95 kg of Pu239, of which 2.5 kg actually undergoes fission. (See Problem 4.) (b) Why is the other 92.5 kg of Pu239needed if it does not fission?

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Find role="math" localid="1661942719139" ∆KKfor each of the following acting as the stationary atom:

(b) hydrogen,

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(e) lead.

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Verify the Q values reported in Eqs. 43-13, 43-14, and 43-15. The needed masses are

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H4e+H4e+H4e→12C+7.27MeV

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