Nuclear Chemistry Worksheet

Nuclear Chemistry Worksheet222 41 86rn 2he234

Complete the following nuclear reactions and decay equations by writing the correct balanced nuclear equations. Include the proper symbols for particles (alpha, beta, gamma) and identify the changes in atomic and mass numbers as appropriate.

For each decay process, write the balanced nuclear equation. Specifically, for alpha decays, include the emission of a helium-4 nucleus; for beta decays, include the emission of a beta particle (electron); and for gamma decay, note the emission of a gamma photon. Indicate how the mass number (A) and atomic number (Z) of the nucleus are affected in each case.

Paper For Above instruction

Let's analyze the given nuclear reactions systematically, completing each by applying principles of nuclear physics. The key to correctly balancing nuclear equations involves preserving both the total atomic number and the total mass number on each side of the equation. The atomic number (Z) defines the element, while the mass number (A) is the total number of protons and neutrons.

Part 1: Completes the given unbalanced reactions

1. The reaction: 86Rn ---------> ________ + 2He

This is an alpha decay of radon-86. An alpha particle (2He or helium-4 nucleus) is emitted. Balance by subtracting the alpha particle's nucleus from radon:

• Atomic number: 86 (radon) - 2 (helium) = 84
• Mass number: 86 - 4 = 82

Thus, the product is polonium-82 (or element with atomic number 84). The balanced equation is:

86Rn → 82Po + 4He

2. The reaction: ________ ---------> 90Th + 2He

This appears to be an alpha decay of thorium-90, producing an alpha particle and a new element. The alpha particle has atomic number 2, so the element produced is:

• Atomic number: 90 (Th) - 2 = 88
• Mass number: 90 - 4 = 86

Element 88 is radium (Ra). The balanced equation is:

90Th → 86Ra + 4He

3. The reaction: 90Th ---------> 88Ra + ______

This is likely the alpha decay of thorium-90 producing radium-88.

The missing particle is again an alpha particle (4He), verifying the previous calculation:

90Th → 88Ra + 4He

4. The reaction: 82Pb ---------> ______ + -1e

This involves beta decay of lead-82. In beta decay, a neutron converts into a proton, emitting a beta particle (electron, -1e). The atomic number increases by 1, but the mass number stays the same:

• Atomic number: 82 (Pb) + 1 = 83
• Mass number: 82 (unaffected)

Element with atomic number 83 is bismuth (Bi). The reaction is:

82Pb → 83Bi + -1e

5. The reaction: 92U ---------> 93Np + _______

This exhibits beta decay of uranium-92. The nucleus gains a proton, increasing atomic number by 1, with the mass number unchanged:

• Atomic number: 92 (U) + 1 = 93
• Mass number: 92

The product is neptunium-93:

92U → 93Np + -1e

Part 2: Write Decay Equations for Specific Elements

a) Nitrogen-16 Beta Decay

Nitrogen-16 undergoes beta decay, converting a neutron into a proton and emitting a beta particle. The equations are:

16^N → 16^O + -1e

In this case, the atomic number increases from 7 (N) to 8 (O), and the mass number remains 16.

b) Potassium-40 Beta Decay

Similarly, potassium-40 decays via beta emission, transforming into calcium-40:

40^K → 40^Ca + -1e

Part 3: Decay of Radium-226

Radium-226 undergoes alpha, beta, and gamma decay steps. Each step involves different particle emissions affecting atomic and mass numbers differently.

Alpha Decay:

226^Ra → 222^Rn + 4He

It decreases the mass number by 4 and the atomic number by 2.

Beta Decay:

Radium-226 can undergo beta decay to produce actinium-226:

226^Ra → 226^Ac + -1e

Atomic number increases by 1; mass number remains unchanged.

Gamma Decay:

Gamma decay involves the emission of gamma radiation, which releases energy but does not change atomic or mass numbers:

ΔEnergy transfer; No change in atomic/mass numbers

Effects on Mass and Atomic Numbers from Particle Loss

• Beta particle: Loss of a beta particle increases the atomic number by 1; mass number remains unchanged.
• Alpha particle: Loss of an alpha particle decreases the mass number by 4 and the atomic number by 2.
• Gamma particle: Gamma emission does not affect mass or atomic number; energy is released without changing the nucleus.

Conclusion

Understanding nuclear decay processes involves recognizing particle emissions and their impact on nuclear composition. Alpha decay reduces mass and atomic numbers significantly, beta decay increases the atomic number while leaving the mass number unchanged, and gamma decay involves pure energy release with no change in nucleus composition. These principles underpin nuclear physics and are critical for applications such as radiometric dating, nuclear medicine, and nuclear power generation.

References

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