Chapter 19, Problem 19.103QP

(a)

Interpretation Introduction

Interpretation:

Equation for the given decay process has to be written.

Concept Introduction:

The total number of protons and neutrons in the product and in reactant side must be equal.

The total number of nuclear charges in the product and reactant side must be equal.

${}_Z^{A}X$, where X is the element, A is the mass number and Z is the atomic number.

Explanation of Solution

When $\text{Po-210}$ isotope decays by bombarding ${}^{209}Bi$ with neutron. The balanced equation is

${}_{83}^{209}Bi+{}_0^{1}n{\to }_{83}^{210}Bi{\to }_{84}^{210}Po+{}_{-1}^0\beta$

(b)

Interpretation Introduction

Interpretation:

The discoverer of element polonium has to be mentioned.

Concept Introduction:

${}_{84}^{210}Po$ has mass number 210 and atomic number 84.

Polonium is found in uranium ore, and is highly radioactive

Explanation of Solution

The stable isotope ${}_{84}^{210}Po$ was discovered by Marie Curie.

(c)

Interpretation Introduction

Interpretation:

Equation for decay process of ${}_{84}^{210}Po$ when decays with the emission of a $\alpha$ particle has to be written.

Concept Introduction:

• Half-life of the reaction is given by

$t_{1/2}=\frac{0.693}{\lambda }$

• The total number of protons and neutrons in the product and in reactant side must be equal.
• The total number of nuclear charges in the product and reactant side must be equal.
• ${}_Z^{A}X$, where X is the element, A is the mass number and Z is the atomic number.

Explanation of Solution

Given,

$t_{1/2}of{}_{84}^{210}Po=138days$

The equation for decay process of ${}_{84}^{210}Po$ when decays with the emission of an $\alpha$ particle is given below,

${}_{84}^{210}Po\to {}_{82}^{206}Po+{}_2^4\alpha$

(d)

Interpretation Introduction

Interpretation:

The energy of an emitted $\alpha$ particle has to be determined.

Concept Introduction:

• Energy required to break the nucleus into its corresponding proton and neutron is called nuclear binding energy
• This quantity represents the conversion of mass to energy occurs during an exothermic reaction.
• Nuclear binding energy can be calculated by Einstein’s mass energy equivalence relationship that is,$\Delta E=\left(\Delta m\right)c^2$

  Where, $\left(\Delta m\right)$ is called mass defect.

• $\text{Δm}\text{=}\text{mass}\text{of}\text{product}\text{-mass}\text{of}\text{reactant}$

Explanation of Solution

The equation for decay process of ${}_{84}^{210}Po$ when decays with the emission of an $\alpha$ particle is given below,

${}_{84}^{210}Po\to {}_{82}^{206}Po+{}_2^4\alpha$

$\begin{array}{l}\text{Mass}\text{of}{}_{\text{82}}^{\text{206}}\text{Po}\text{=}\text{205}\text{.9744amu}\\ \text{Mass}\text{of}\text{α}\text{particle=4}\text{.00150}\text{amu}\\ \text{Total}\text{mass}\text{of}\text{product}\text{=205}\text{.97444amu+4}\text{.00150}\text{amu=209}\text{.97594amu}\\ \\ \text{Mass}\text{of}{}_{\text{84}}^{\text{210}}\text{Po}\text{=209}\text{.98285amu}\\ \text{Total}\text{mass}\text{of}\text{reactant}\text{=209}\text{.98285amu}\\ \\ \text{Δm}\text{=}\text{mass}\text{of}\text{product}\text{-mass}\text{of}\text{reactant}\\ \text{Δm}\text{=209}\text{.97594amu-209}\text{.98285amu}\\ \text{Δm}\text{=-6}{\text{.91×10}}^{\text{-3}}\text{amu×}\frac{\text{1kg}}{\text{6}{\text{.023×10}}^{\text{26}}\text{amu}}\text{=-1}{\text{.147×10}}^{\text{-29}}\text{kg}\end{array}$

$\begin{array}{l}\text{ΔE =}\left(\text{Δm}\right){\text{c}}^{\text{2}}\\ \text{ΔE =-1}{\text{.147×10}}^{\text{-29}}\text{kg×}{\left({\text{3×10}}^{\text{8}}\text{m/s}\right)}^{\text{2}}\\ \text{ΔE =-1}\text{.032}\times {10}^{-12}\text{kg}{\text{m}}^{\text{2}}{\text{/s}}^{\text{2}}\\ \text{ΔE =-1}\text{.032}\times {10}^{-12}\text{J}\end{array}$

The binding energy of a single neutron is $\text{1}\text{.032}\times {10}^{-12}\text{J}$

(e)

Interpretation Introduction

Interpretation:

The total energy releases by ${}_{84}^{210}Po$ has to be determined.

Concept Introduction:

• Energy required to break the nucleus into its corresponding proton and neutron is called nuclear binding energy
• This quantity represents the conversion of mass to energy occurs during an exothermic reaction.
• Nuclear binding energy can be calculated by Einstein’s mass energy equivalence relationship that is,$\Delta E=\left(\Delta m\right)c^2$

  Where, $\left(\Delta m\right)$ is called mass defect.

• $\text{Δm}\text{=}\text{mass}\text{of}\text{product}\text{-mass}\text{of}\text{reactant}$

Explanation of Solution

The binding energy of a single neutron is $\text{1}\text{.032}\times {10}^{-12}\text{J}$

Given,

Mass $=1\mu g=1\times {10}^{-6}g$

The total energy releases by ${}_{84}^{210}Po$ can be calculated as given below,

$1\times {10}^{-6}g{}_{84}^{210}Po\times \frac{1mol{}_{84}^{210}Po}{209.98285{}_{84}^{210}Po}\times \frac{1mol\alpha }{1mol\alpha }\times \frac{6.022\times {10}^{23}}{1mol\alpha }\times \frac{\text{1}\text{.032}\times {10}^{-12}\text{J}}{1\alpha particle}=2.959\times {10}^{3}J$