Question 43.7: The Energy Liberated When Radium Decays The ^226Ra nucleus u...

The Energy Liberated When Radium Decays

The { }^{226} Ra nucleus undergoes alpha decay according to Equation 43.12.

{ }_{88}^{226}Ra \rightarrow{ }_{86}^{222} Rn+{ }_2^4 He     (43.12)

(A) Calculate the Q value for this process. From Table 43.2, the masses are 226.025 408 u for { }^{226} Ra, 222.017 576 u for { }^{222} Rn, and 4.002 603 u for { }_2^4 He.

(B) What is the kinetic energy of the alpha particle after the decay?

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(A) Conceptualize Study Figure 43.11 to understand the process of alpha decay in this nucleus.

Categorize The parent nucleus is an isolated system that decays into an alpha particle and a daughter nucleus. The system is isolated in terms of both energy and momentum.

Analyze Evaluate Q using Equation 43.15:

\begin{aligned}Q & =\left(M_{X}-M_{Y}-M_\alpha\right) \times 931.494  MeV / u \\& =(226.025  408  u-222.017  576  u-4.002  603  u) \times 931.494  MeV / u\\& =(0.005  229  u) \times 931.494  MeV / u=4.87  MeV\end{aligned}

(B) Analyze The value of 4.87 MeV is the disintegration energy for the decay. It includes the kinetic energy of both the alpha particle and the daughter nucleus after the decay. Therefore, the kinetic energy of the alpha particle would be less than 4.87 MeV.

Set up a conservation of momentum equation, noting that the initial momentum of the system is zero:

(1)   0=M_{Y} v_{Y}-M_\alpha v_\alpha

Solve Equation 43.13 for the negative of the change in the  rest mass and then express the left side of the equation as Q using Q=-\Delta E_R. On the right side, express the final kinetic energy of the system as the sum of kinetic energies of the daughter nucleus and the alpha particle:

\Delta E_R+\Delta K=0     (43.13)

-\Delta E_R=\Delta K \rightarrow Q=\Delta K

(2)   Q=\frac{1}{2} M_\alpha v_\alpha{}^2+\frac{1}{2} M_{Y} v_{Y}{}^2

Solve Equation (1) for v_{Y} and substitute into Equation (2). Solve the result for the kinetic energy of the alpha particle:

\begin{aligned}Q & =\frac{1}{2} M_\alpha v_\alpha{}^2+\frac{1}{2} M_{Y}\left(\frac{M_\alpha v_\alpha}{M_{Y}}\right)^2=\frac{1}{2} M_\alpha v_\alpha{}^2\left(1+\frac{M_\alpha}{M_{Y}}\right) \\& =K_\alpha\left(\frac{M_{Y}+M_\alpha}{M_{Y}}\right) \rightarrow K_\alpha=Q\left(\frac{M_{Y}}{M_{Y}+M_\alpha}\right)\end{aligned}

Evaluate this kinetic energy for the specific decay of { }^{226} Ra that we are exploring in this example:

K_\alpha=(4.87  MeV)\left(\frac{222}{222+4}\right)=4.78  MeV

Finalize The kinetic energy of the alpha particle is indeed less than the disintegration energy, but notice that the alpha particle carries away most of the energy available in the decay.

43.7

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