In the liquid-drop model, the mass of a nucleus with mass number \(A\) can be expressed as a quadratic in \(Z: M(A, Z)=\) \(c_{1} A-c_{2} Z+\left(c_{2} A^{-1}+c_{3} A^{-1 / 3}\right) Z^{2},\) where the \(c s\) are constants determined from experimental data. Show that the value of \(Z\) that gives the minimum mass (not necessarily an integer) is \(Z_{\min }=\) \((A / 2) /\left[1+\left(c_{3} / c_{2}\right) A^{2 / 3}\right] .\) (Note: A plot of \(Z_{\min }\) versus \(N=A-Z\) gives the line of greatest nuclear stability in Figs. 38.3 and \(38.12 .\) )

Short Answer

Expert verified
The minimum mass in the liquid-drop model is acquired when \(Z = A/2[1 + (c_{3}/c_{2})A^{2/3}]\)

Step by step solution

01

Understanding the Function

The given function is \(M(A, Z)= c_{1} A-c_{2} Z+(c_{2} A^{-1}+c_{3} A^{-1 / 3}) Z^{2}\) which models the mass of a nucleus. In order to find the minimum mass, you will have to find the value of \(Z\) that minimizes this function.
02

Taking the Derivative

The first step to find the point at which this function reaches a minimum is to take derivative of \(M\) with respect to \(Z\) and set it equal to zero. After differentiating, you'll get \(-c_{2} + 2Z(c_{2}A^{-1} + c_{3}A^{-1/3}) = 0\)
03

Solving for Z

Solving the above equation for \(Z\), you get \(Z_{\min} = c_{2}/\left[2(c_{2}A^{-1} + c_{3}A^{-1/3})\right]\)
04

Simplifying the Equation

The numerator can be simplified as \(Z_{\min} = A/2[1 + (c_{3}/c_{2})A^{2/3}]\) giving the minimum value of Z.

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Most popular questions from this chapter

Passage Problems In \(1972,\) a worker at a nuclear fuel plant in France discovered that uranium from a mine in Oklo, in the African Republic of Gabon, had less U-235 than the normal 0.7 \(\%\) - a quantity known from meteorites and Moon rocks to be constant throughout the solar system. Further analysis showed the presence of isotopes that would result from the decay of fission products. Scientists drew the remarkable conclusion that a natural nuclear fission reaction had occurred some 2 billion years ago, lasting for about 100,000 years. Water, mixing with rich uranium ore, provided the moderator that made the chain reaction possible. More significantly, U-235's 700-million-year half-life means that 2 billion years ago there was a higher abundance of \(U-235\) in natural uranium. At the time of the Oklo fission reaction, the actual amount of U-235 present was a. about the same as today. b. about twice as much as today. c. about four times as much as today. d. about eight times as much as today.

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