Chapter 11, Problem 11.44EP

(a)

Interpretation Introduction

Interpretation:

Equation for nickel-58 bombarded with an proton giving alpha particle as product has to be written.

Concept Introduction:

Radioactive nuclides undergo disintegration by emission of radiation.  This is a natural transmutation reaction where the nuclide of one element is converted into nuclide of another element.  Radioactive decay happens naturally.  This can also be done artificially in the laboratory by means of bombardment reaction.  Bombardment reaction is the one where the target nuclei is hit by a small fast moving high-energy particle to give a daughter nuclide and a small particle such as proton or neutron.  This can be represented in form of a nuclear equation.  A balanced nuclear equation is the one in which the sum of subscripts on both sides are equal and sum of superscripts on both sides are equal.

Answer to Problem 11.44EP

The equation is,

    $\text{N}\text{i}+\text{H}\stackrel{}{\to }\text{C}\text{o}+\alpha$

Explanation of Solution

Given description is an alpha particle bombards with beryllium-9 to give a neutron product.  The incomplete equation for this can be written as,

$\text{N}\text{i}+\text{H}\stackrel{}{\to }?+\alpha$

In a nuclear equation the sum of subscript on both sides has to be equal and the sum of superscripts on both sides has to be equal.  By looking into the above equation, the sum of superscript in the product side is 4 and the sum of subscript in the product side is 2.  This has to be balanced in the reactant side.  Sum of superscript in the reactant side is 59.  Sum of subscript is 29.  There is a short of 55 in the superscript and 27 in the subscript of the product side when comparing with the reactant side.  The element with atomic number 27 is cobalt.  Therefore, the missing nuclear symbol is $\text{C}\text{o}$.  The complete nuclear equation can be given as,

$\text{N}\text{i}+\text{H}\stackrel{}{\to }\text{C}\text{o}+\alpha$

Conclusion

The equation for proton bombarding with nickel-58 is written.

(b)

Interpretation Introduction

Interpretation:

Equation for aluminium-27 bombarded with an alpha particle giving neutron as product has to be written.

Concept Introduction:

Radioactive nuclides undergo disintegration by emission of radiation.  This is a natural transmutation reaction where the nuclide of one element is converted into nuclide of another element.  Radioactive decay happens naturally.  This can also be done artificially in the laboratory by means of bombardment reaction.  Bombardment reaction is the one where the target nuclei is hit by a small fast moving high-energy particle to give a daughter nuclide and a small particle such as proton or neutron.  This can be represented in form of a nuclear equation.  A balanced nuclear equation is the one in which the sum of subscripts on both sides are equal and sum of superscripts on both sides are equal.

Answer to Problem 11.44EP

The equation is,

    $\text{A}\text{l}+\alpha \stackrel{}{\to }\text{P}+\text{n}$

Explanation of Solution

Given description is an alpha particle bombards with aluminium-27 to give a neutron product.  The incomplete equation for this can be written as,

$\text{A}\text{l}+\alpha \stackrel{}{\to }?+\text{n}$

In a nuclear equation the sum of subscript on both sides has to be equal and the sum of superscripts on both sides has to be equal.  By looking into the above equation, the sum of superscript in the product side is 1 and the sum of subscript in the product side is 0.  This has to be balanced in the reactant side.  Sum of superscript in the reactant side is 31.  Sum of subscript is 15.  There is a short of 30 in the superscript and 15 in the subscript of the product side when comparing with the reactant side.  The element with atomic number 15 is phosphorus.  Therefore, the missing nuclear symbol is $\text{P}$.  The complete nuclear equation can be given as,

$\text{A}\text{l}+\alpha \stackrel{}{\to }\text{P}+\text{n}$

Conclusion

The equation for alpha particle bombarding with aluminium-27 is written.

(c)

Interpretation Introduction

Interpretation:

Equation for curium-246 giving nobelium-254 and four neutrons on bombardment with a small particle has to be written.

Concept Introduction:

Radioactive nuclides undergo disintegration by emission of radiation.  This is a natural transmutation reaction where the nuclide of one element is converted into nuclide of another element.  Radioactive decay happens naturally.  This can also be done artificially in the laboratory by means of bombardment reaction.  Bombardment reaction is the one where the target nuclei is hit by a small fast moving high-energy particle to give a daughter nuclide and a small particle such as proton or neutron.  This can be represented in form of a nuclear equation.  A balanced nuclear equation is the one in which the sum of subscripts on both sides are equal and sum of superscripts on both sides are equal.

Answer to Problem 11.44EP

The equation is,

    $\text{C}\text{m}+\text{C}\stackrel{}{\to }\text{N}\text{o}+4\text{n}$

Explanation of Solution

Given description is a small particle bombards with curium-246 to give nobelium-254 and four neutrons as product.  The incomplete equation for this can be written as,

$\text{C}\text{m}+?\stackrel{}{\to }\text{N}\text{o}+4\text{n}$

In a nuclear equation the sum of subscript on both sides has to be equal and the sum of superscripts on both sides has to be equal.  By looking into the above equation, the sum of superscript in the product side is 258 and the sum of subscript in the product side is 102.  This has to be balanced in the reactant side.  Sum of superscript in the reactant side is 246.  Sum of subscript is 96.  There is a short of 12 in the superscript and 6 in subscript of the reactant side when comparing with the product side.  The element with the atomic number 6 is carbon.  Therefore, the missing nuclear symbol is $\text{C}$.  The complete nuclear equation can be given as,

$\text{C}\text{m}+\text{C}\stackrel{}{\to }\text{N}\text{o}+4\text{n}$

Conclusion

The equation for curium-246 converting into nobelium-254 and four neutrons by a small particle by bombarding is written.

(d)

Interpretation Introduction

Interpretation:

Equation for bombarding a nuclide with alpha particle that gives curium-252 and neutron has to be written.

Concept Introduction:

Radioactive nuclides undergo disintegration by emission of radiation.  This is a natural transmutation reaction where the nuclide of one element is converted into nuclide of another element.  Radioactive decay happens naturally.  This can also be done artificially in the laboratory by means of bombardment reaction.  Bombardment reaction is the one where the target nuclei is hit by a small fast moving high-energy particle to give a daughter nuclide and a small particle such as proton or neutron.  This can be represented in form of a nuclear equation.  A balanced nuclear equation is the one in which the sum of subscripts on both sides are equal and sum of superscripts on both sides are equal.

Answer to Problem 11.44EP

The equation is,

    $\text{P}\text{u}+\alpha \stackrel{}{\to }\text{C}\text{m}+\text{n}$

Explanation of Solution

Given description is a nuclide is bombarded by alpha particle that gives curium-252 and neutron.  The incomplete equation for this can be written as,

$?+\alpha \stackrel{}{\to }\text{C}\text{m}+\text{n}$

In a nuclear equation the sum of subscript on both sides has to be equal and the sum of superscripts on both sides has to be equal.  By looking into the above equation, the sum of superscript in the product side is 253 and the sum of subscript in the product side is 98.  This has to be balanced in the reactant side.  Sum of superscript in the reactant side is 4.  Sum of subscript is 2.  There is a short of 249 in the superscript and a short of 96 in subscript of the reactant side when comparing with the product side.  Plutonium is the element that has atomic number 96.  Therefore, the missing nuclear symbol is $\text{P}\text{u}$.  The complete nuclear equation can be given as,

$\text{P}\text{u}+\alpha \stackrel{}{\to }\text{C}\text{m}+\text{n}$

Conclusion

The equation for alpha particle bombarding with a nuclide giving curium-252 and neutron is given.