/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 51 The superheavy element \(^{287} ... [FREE SOLUTION] | 91Ó°ÊÓ

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The superheavy element \(^{287} \mathrm{Fl}\) (element 114 ) was made by firing a beam of \(^{48}\) Ca ions at \(^{242}\) Pu. Three neutrons were ejected in the reaction. Write a balanced nuclear equation for the synthesis of \(^{287} \mathrm{Fl}\).

Short Answer

Expert verified
The balanced equation is \( ^{48}_{20}\mathrm{Ca} + ^{242}_{94}\mathrm{Pu} \rightarrow ^{287}_{114}\mathrm{Fl} + 3^1_0\mathrm{n} \).

Step by step solution

01

Identify Reactants and Products

The reactants in the nuclear reaction are the calcium isotope \(^{48}\)Ca and the plutonium isotope \(^{242}\)Pu. The product of the reaction is the superheavy element flerovium \(^{287}\)Fl. Additionally, three neutrons (\(^1_0n\)) are ejected as part of this reaction.
02

Write the Nuclear Reaction

In a nuclear reaction, the sum of the mass numbers and the atomic numbers needs to be conserved. The equation for the reaction can be written as follows: \[ ^{48}_{20}\mathrm{Ca} + ^{242}_{94}\mathrm{Pu} \rightarrow ^{287}_{114}\mathrm{Fl} + 3^1_0\mathrm{n} \]
03

Verify Mass and Atomic Number Conservation

Check the conservation of mass numbers and atomic numbers on both sides of the equation:- The sum of mass numbers on the left is \(48 + 242 = 290\) and on the right is \(287 + (3 \times 1) = 290\).- The sum of atomic numbers on the left is \(20 + 94 = 114\) and on the right is \(114 + 0 = 114\).This confirms both mass and atomic number conservation in the balanced equation.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Superheavy Elements
Superheavy elements are fascinating substances that reside at the far edge of the periodic table. These elements, like flerovium (Fl), are not naturally occurring and are produced through complex nuclear reactions. Flerovium, also known as element 114, was first synthesized by scientists who aimed to discover new elements by colliding smaller atomic nuclei together. This process involves using heavy ion beams, such as calcium ions, to bombard target materials like plutonium. The result is the creation of elements that possess incredibly high atomic numbers.

Creating superheavy elements is complex due to their instability. They often exist for very short periods before decaying into other elements. Despite their fleeting nature, studying these elements helps scientists understand the forces that bind atomic nuclei and explore theoretical predictions about the "island of stability," where certain superheavy isotopes might have relatively longer lifetimes.
Nuclear Reactions
Nuclear reactions involve changes in the nucleus of atoms, different from chemical reactions which involve electrons. In nuclear reactions, elements can transform into different elements, as seen in the synthesis of superheavy elements like flerovium (Fl).

During these reactions, the atomic nuclei collide and combine to form a new nucleus. In the reaction involving calcium-48 ions and plutonium-242, the nuclei fused to form flerovium-287. This process typically requires careful control of the reaction environment to achieve.

At the heart of nuclear reactions are key principles such as conservation laws. In our example, it's crucial to control the reaction energy and particle velocities to manage the formation of desired elements efficiently without unnecessary byproducts.
Mass Number Conservation
Mass number conservation is a fundamental principle in nuclear chemistry. It states that the sum of mass numbers (protons plus neutrons) in the reactants must equal the sum in the products in a nuclear reaction.

In the synthesis of flerovium-287 ( Fl ), the mass numbers add up perfectly: the calcium-48 and plutonium-242 react to give flerovium-287 and three neutrons. Here's how it works:

  • The sum of the mass numbers of reactants: 48 (from Ca) + 242 (from Pu) = 290.
  • The sum of the mass numbers of products: 287 (from Fl) + 3 (from three neutrons each having a mass number of 1) = 290.

This confirms the conservation of mass numbers. Similarly, atomic number conservation is also crucial to ensure that the sum of atomic numbers remains constant during the nuclear reaction. This symmetry helps verify the feasibility and success of the equation.

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Most popular questions from this chapter

Collision of an electron and a positron results in formation of two \(\gamma\) rays. In the process, their masses are converted completely into energy. (a) Calculate the energy evolved from the annihilation of an electron and a positron, in kilojoules per mole. (b) Using Planck's equation (Equation 6.2), determine the frequency of the \(\gamma\) rays emitted in this process.

Phosphorus-32 is used in the form of \(\mathrm{Na}_{2} \mathrm{HPO}_{4}\) in the treatment of chronic myeloid leukemia, among other things. (a) The isotope decays by \(\beta\) -particle emission. Write a balanced equation for this process. (b) The half-life of \(^{32} \mathrm{P}\) is 14.3 days. If you begin with \(4.8 \mu \mathrm{g}\) of radioactive \(^{32} \mathrm{P}\) in the form of \(\mathrm{Na}_{2} \mathrm{HPO}_{4},\) what mass remains after 28.6 days (about 1 month)?

Predict the probable mode of decay for each of the following radioactive isotopes, and write an equation to show the products of decay. (a) manganese- 54 (b) americium- 241 (c) silver-110 (d) mercury-197m

To synthesize the heavier transuranium elements, a nucleus must be bombarded with a relatively large particle. If you know the products are californium- 246 and four neutrons, with what particle would you bombard uranium- 238 atoms?

Some of the reactions explored by Ernest Rutherford (pages 67 and 1166 ) and others are listed below. Identify the unknown species in each reaction. (a) \(^{14} \mathrm{N}+_{2}^{4} \mathrm{He} \longrightarrow^{17}_{8} \mathrm{O}+?\) (b) \(^{9}_{4} \mathrm{Be}+_{2}^{4} \mathrm{He} \longrightarrow ?+\\_{1}^{6} \mathrm{n}\) (c) \(?+\\_{4}^{2} \mathrm{He} \longrightarrow_{15}^{30} \mathrm{P}+\\_{1}^{0} \mathrm{n}\) (d) \(^{233} \mathrm{Pu}+_{2}^{4} \mathrm{He} \longrightarrow ?+\\_{1}^{0} \mathrm{n}\)

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