Chapter 42: Q 5 Conceptual Questions (page 1235)

Are the following decays possible? If not, why not?
$a. {}^{232}{Th} (Z = 90) \to {}^{236}U (Z = 92) + α b. {}^{238}{Pu} (Z = 94) \to {}^{236}U (Z = 92) + α c. {}^{11}B (Z = 5) \to {}^{11}B (Z = 5) + γ d. {}^{33}P (Z = 15) \to {}^{32}S (Z = 16) + e^{-}$

Short Answer

Expert verified

Therefore, the decays are explained.

Step by step solution

Given information

The following decays are given:

a. {}^{232}{Th} (Z = 90) \to {}^{236}U (Z = 92) + α b. {}^{238}{Pu} (Z = 94) \to {}^{236}U (Z = 92) + α c. {}^{11}B (Z = 5) \to {}^{11}B (Z = 5) + γ d. {}^{33}P (Z = 15) \to {}^{32}S (Z = 16) + e^{-}

Explanation (a)

(a)No, the alpha particle is inserted into the incorrect nucleus.

Explanation (b)

(b)No, the original nucleus would have to lose four nucleons rather than two.

Explanation (c)

(c)Yes.

Explanation (d)

(d) No, A does not change during beta decay.

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Are the following decays possible? If not, why not?
$a. {}^{232}{Th} (Z = 90) \to {}^{236}U (Z = 92) + α b. {}^{238}{Pu} (Z = 94) \to {}^{236}U (Z = 92) + α c. {}^{11}B (Z = 5) \to {}^{11}B (Z = 5) + γ d. {}^{33}P (Z = 15) \to {}^{32}S (Z = 16) + e^{-}$

Short Answer

Step by step solution

Given information

Explanation (a)

Explanation (b)

Explanation (c)

Explanation (d)

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Are the following decays possible? If not, why not?a.Th232(Z=90)→U236(Z=92)+αb.Pu238(Z=94)→U236(Z=92)+αc.B11(Z=5)→B11(Z=5)+γd.P33(Z=15)→S32(Z=16)+e-

Short Answer

Step by step solution

Given information

Explanation (a)

Explanation (b)

Explanation (c)

Explanation (d)

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Most popular questions from this chapter

Recommended explanations on Physics Textbooks

Electricity

Fluids

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Engineering Physics

Study anywhere. Anytime. Across all devices.

Are the following decays possible? If not, why not?
$a. {}^{232}{Th} (Z = 90) \to {}^{236}U (Z = 92) + α b. {}^{238}{Pu} (Z = 94) \to {}^{236}U (Z = 92) + α c. {}^{11}B (Z = 5) \to {}^{11}B (Z = 5) + γ d. {}^{33}P (Z = 15) \to {}^{32}S (Z = 16) + e^{-}$