Chapter 19, Problem 19.7QP

(a)

Interpretation Introduction

Interpretation: Determine the X nuclear species for the following nuclear equation.

Concept Introduction:

In balancing any nuclear equation, we must balance the total of all atomic numbers and the total of all mass. If we know the atomic numbers and mass numbers of all but one of the species in a nuclear equation

The alpha (${}_{\text{2}}^{\text{4}}\text{α}$) –particle has two protons and two neutrons. So its atomic number is 2 and its mass number is 4.

For a beta- particle ($\text{β}{}^{\text{0}}$) is the negative unit charge, which atomic number is decreased by one unit and no change in atomic mass.

For neutron particle (${}_{\text{0}}^{\text{1}}\text{n}$) no change in the atomic number only the atomic mass increased by one unit.

To Determine: Identify the value of X.

Answer to Problem 19.7QP

${}_{\text{12}}^{\text{26}}\text{Mg+}\text{P}{}_{\text{1}}^{\text{1}}\stackrel{}{\to }\text{α+}{\text{Na}}_{\text{11}}^{\text{23}}$

Explanation of Solution

In balancing nuclear equation, the sum of atomic number and the mass number must match on both sides of the equation.

 On the left side of the equation, atomic number sum is 13(12+1) and the mass number sum is 27(26+1).these sum must be the same on the right side of the same equation.

Note that the atomic and mass number of an alpha particle are 2and 4, respectively.

For a beta- particle ($\text{β}{}^{\text{0}}$) is the negative unit of atomic number and the mass no is same (0) adding nothing to atomic mass.

For neutron particle (${}_{\text{0}}^{\text{1}}\text{n}$) no change in the atomic number only the atomic mass contain by the neutron particle is added or accounted.

So the number of X is there for 11(13-2) and the mass number is 23(27-4).

So the X is sodium -23(${\text{Na}}_{\text{11}}^{\text{23}}$).

(b)

Interpretation Introduction

Interpretation: Determine the X nuclear species for the following nuclear equation.

Concept Introduction:

In balancing any nuclear equation, we must balance the total of all atomic numbers and the total of all mass. If we know the atomic numbers and mass numbers of all but one of the species in a nuclear equation

The alpha (${}_{\text{2}}^{\text{4}}\text{α}$) –particle has two protons and two neutrons. So its atomic number is 2 and its mass number is 4.

For a beta- particle ($\text{β}{}^{\text{0}}$) is the negative unit charge, which atomic number is decreased by one unit and no change in atomic mass.

For neutron particle (${}_{\text{0}}^{\text{1}}\text{n}$) no change in the atomic number only the atomic mass increased by one unit.

To Determine: Identify the value of X.

Answer to Problem 19.7QP

${\text{Co}}_{\text{27}}^{\text{59}}\text{+}{\text{H}}_{\text{1}}^{\text{2}}\stackrel{}{\to }{\text{CO}}_{\text{27}}^{\text{60}}\text{+}{\text{H}}_{\text{1}}^{\text{1}}$

Explanation of Solution

On the left side of the equation the atomic number sum is 28(27+1) and the mass number sum is 61(59+2).

These sums must be the same on the right side.

Note that the atomic and mass number of an alpha particle are 2and 4, respectively.

For a beta- particle ($\text{β}{}^{\text{0}}$) is the negative unit of atomic number and the mass no is same (0) adding nothing to atomic mass.

For neutron particle (${}_{\text{0}}^{\text{1}}\text{n}$) no change in the atomic number only the atomic mass contain by the neutron particle is added or accounted.

The atomic number of X is there1 (13-2).and the mass number is also 1(61-60).

X is a proton (${\text{H}}_{\text{1}}^{\text{1}}$).

(c)

Interpretation Introduction

Interpretation: Determine the X nuclear species for the following nuclear equation.

Concept Introduction:

In balancing any nuclear equation, we must balance the total of all atomic numbers and the total of all mass. If we know the atomic numbers and mass numbers of all but one of the species in a nuclear equation

The alpha (${}_{\text{2}}^{\text{4}}\text{α}$) –particle has two protons and two neutrons. So its atomic number is 2 and its mass number is 4.

For a beta- particle ($\text{β}{}^{\text{0}}$) is the negative unit charge, which atomic number is decreased by one unit and no change in atomic mass.

For neutron particle (${}_{\text{0}}^{\text{1}}\text{n}$) no change in the atomic number only the atomic mass increased by one unit.

To Determine: Identify the value of X.

Answer to Problem 19.7QP

${}_{\text{92}}^{\text{235}}\text{U}\text{+}{}_{\text{0}}^{\text{1}}\text{n}\stackrel{}{\to }{}_{\text{36}}^{\text{94}}\text{Kr}\text{+}{}_{\text{56}}^{\text{139}}{\text{Ba+3}}_{\text{0}}^{\text{1}}\text{n}$

Explanation of Solution

In balancing nuclear equation, the sum of atomic number and the mass number must match on both sides of the equation.

  On the left side of the equation, atomic number sum is 236(235+1) and the mass number sum is 92(92+0).these sum must be the same on the right side of the same equation.

For a beta- particle ($\text{β}{}^{\text{0}}$) is the negative unit of atomic number and the mass no is same (0) adding nothing to atomic mass.

For neutron particle (${}_{\text{0}}^{\text{1}}\text{n}$) no change in the atomic number only the atomic mass contain by the neutron particle is added or accounted.

On the right side of the equation, atomic number sum is 233 = (94+139) and the mass number sum is 92(92+0). These sum must be the same on the right side of the same equation.

So the number of X is there for mass number (3) and the 0 for atomic number is 3(3*1)

So the X is neutron ${}_{\text{0}}^{\text{1}}\text{n}$

 (d)

Interpretation Introduction

Interpretation: Determine the X nuclear species for the following nuclear equation.

Concept Introduction:

In balancing any nuclear equation, we must balance the total of all atomic numbers and the total of all mass. If we know the atomic numbers and mass numbers of all but one of the species in a nuclear equation

The alpha (${}_{\text{2}}^{\text{4}}\text{α}$) –particle has two protons and two neutrons. So its atomic number is 2 and its mass number is 4.

For a beta- particle ($\text{β}{}^{\text{0}}$) is the negative unit charge, which atomic number is decreased by one unit and no change in atomic mass.

For neutron particle (${}_{\text{0}}^{\text{1}}\text{n}$) no change in the atomic number only the atomic mass increased by one unit.

To Determine: Identify the value of X.

Answer to Problem 19.7QP

${}_{\text{24}}^{\text{53}}\text{Cr}\text{+}{}_{\text{2}}^{\text{4}}\text{α}\stackrel{}{\to }{\text{Fe}}_{\text{26}}^{\text{56}}{\text{+}}_{\text{0}}^{\text{1}}\text{n}$

Explanation of Solution

On the left side of the equation the atomic number sum is 92(92/0) and the mass number and the mass number sum is 236(235-1).

These sums must be the same as the sum in right side and the mass number is same on the rigid also.

The atomic number of x there of X is therefore $\left[\frac{92-(36+56)}{3}\right]$ and the mass number1. X is a neutron (${}_{\text{0}}^{\text{1}}\text{n}$).

On the left side of the equation the atomic number sum is 26(24+2) and the mass number sum is 57(53+4).

These sums must be the same on the right side.

Note that the atomic and mass number of an alpha particle are 2unit and 4, respectively.

For a beta- particle ($\text{β}{}^{\text{0}}$) is the negative unit of atomic number and the mass no is same (0) adding nothing to atomic mass.

For neutron particle (${}_{\text{0}}^{\text{1}}\text{n}$) no change in the atomic number only the atomic mass contain by the neutron particle is added or accounted.

The atomic number of X is therefore 26 (26-0).and the mass number is also 56(57-1).

X is iron-56(${}_{\text{26}}^{\text{56}}\text{Fe}$).

(e)

Interpretation Introduction

Interpretation: Determine the X nuclear species for the following nuclear equation.

Concept Introduction:

In balancing any nuclear equation, we must balance the total of all atomic numbers and the total of all mass. If we know the atomic numbers and mass numbers of all but one of the species in a nuclear equation

The alpha (${}_{\text{2}}^{\text{4}}\text{α}$) –particle has two protons and two neutrons. So its atomic number is 2 and its mass number is 4.

For a beta- particle ($\text{β}{}^{\text{0}}$) is the negative unit charge, which atomic number is decreased by one unit and no change in atomic mass.

For neutron particle (${}_{\text{0}}^{\text{1}}\text{n}$) no change in the atomic number only the atomic mass increased by one unit.

To Determine: Identify the value of X.

Answer to Problem 19.7QP

${}_{\text{8}}^{\text{20}}\text{O}\text{+}\stackrel{}{\to }{\text{F}}_{\text{9}}^{\text{20}}{\text{+}}_{\text{-1}}^{\text{0}}\text{β}$

Explanation of Solution

On the left side of the equation atomic number sum is 8 and mass number sum is 20.

The sums must be the same on the right side.

The atomic number of X is there for -1(8-9) and the mass number is 0(20-20).

X is a beta-(${}_{\text{-1}}^{\text{0}}\text{β}$) particle.