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Chapter 11: Q44E (page 519)

Find Q for the decay of beryllium-10

Short Answer

Expert verified

The $Q$ factor is $0.56 MeV$ .

Step by step solution

01

Given data

Given that, $_{4}^{10} Be \to_{5}^{10} B +_{- 1}^{0} e + \bar{v}$ .

Atomic mass of $_{4}^{10} Be, m_{Be} = 10.013534 u$ .

Atomic mass of $_{5}^{10} B, m_{B} = 10.012937 u$ .

02

Definition of Kinetic Energy

The kinetic energy released is, $Q = (m_{i} - m_{f}) c^{2}$ .

Where, $m_{i}$ is the mass of parents’ nucleus, $m_{f}$ is the mass of the daughter nucleus, and $c$ is the speed of light in vacuum.

03

Determine the Kinetic Energy released

Kinetic energy released, $Q = (m_{i} - m_{f}) c^{2}$ .

The conversion factor may be used to express, $c^{2} \to MeV / u$ .

$\begin{matrix} Q = (m_{Be} - m_{B}) (931.5 \frac{MeV}{u}) \\ Q = (10.013534 u - 10.012937 u) (931.5 \frac{MeV}{u}) \\ Q = 0.56 MeV \end{matrix}$

The $Q$ factor is $0.56 MeV$ .

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Most popular questions from this chapter

Exercise 19 notes that the energy needed to remove a neutron fron helium-4 is 20.6 MeV

(a) Show that the energy required to remove a proton is 19.8 MeV.

(b) Why do these values disagree with the BE/nucleon value shown for helium-4 in Figure 11.14?

Oxygen-19 $β^{-}$ decays. What is the daughter nucleus, and what may be said of the kinetic energy of the emitted $β^{-}$ particle?

The vast majority of measured nuclear spins are fairly small, between zero and nine times $\frac{1}{2} h$ . More to the point. Nuclear spin does not uniformly increases with $A$ . Argue that if the assumptions in the shell model are valid then these observations are not unexpected.

Calculate the binding energy per nucleon of technetium - $98$ .

By classical, hard sphere assumption, what smallest value of $A$ would make one nucleon surrounded. Relate your answer with figure 11.14.

See all solutions

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