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hjlf
3 years ago
10

When aluminum-27 is bombarded with a neutron, a gamma ray is emitted. what radioactive isotope is produced?

Physics
2 answers:
alexandr402 [8]3 years ago
8 0

Answer:

Go to the excited aluminum-28 isotope

Explanation:

In the processes of nuclear reactions, fundamental changes occur at the level of the atomic nucleus.

Analyze the characteristics of the neutron that has approximately the mass of the proton and has no electric charge, so when being absorbed by the aluminum core the atomic number should not be changed, but the atomic mass should increase by one unit.

The new atom formed is in an excited state and enters the base state with the emission of a high-energy radiation, gamma ray that does not transform the atom

Tthe reaction is:

     ⁷A₂₇ + ⁰n₁ → ⁷Al₂₈ˣ

The symbol "ˣ" means excited

      ⁷Al₂₈ˣ → ⁷Al₂₈ + γ

I am Lyosha [343]3 years ago
5 0
Aluminium 28 i think thats the best isotope for this

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The equation that defines the linear moment is given by

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V_f = Final Velocity

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Our values are:

m_1=5.3kgm_2=7.9kg\\m_3=10.5kg\\m_A=75kg\\m_{Total}=m=98.7Kg

Solving to find the final speed, after throwing the object we have

V_f=\frac{mV_0+m_TV_O}{m-m_O}

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That way during each section the equations should be modified depending on the previous one, let's start:

A) 5.3Kg\rightarrow 15m/s

V_{f1}=\frac{mV_0+m_TV_O}{m-m_O}

V_{f1}=\frac{(98.7)*0+5.3*15}{98.7-5.3}

V_{f1}=0.8511m/s

B) 7.9Kg\rightarrow 11.2m/s

V_{f2}=\frac{mV_{f1}+m_TV_O}{m-m_O}

V_{f2}=\frac{(98.7)(0.8511)+(7.9)(11.2)}{98.7-5.3-7.9}

V_{f2} = 2.0173m/s

C) 10.5Kg\rightarrow 7m/s

V_{f3}=\frac{mV_{f2}+m_TV_O}{m-m_O}

V_{f3}=\frac{(98.7)(2.0173)+(10.5)(7)}{98.7-5.3-7.9-10.5}

V_{f3} = 3.63478m/s

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3 years ago
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Two points charges are brought closer together,increasing the force between them by a factor of 25.By what factor wa their separ
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Answer:

The separation between the charges was decreased by a factor of 0.2

Explanation:

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Simora [160]

Answer:

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