Chapter 3, Problem 3.8QP

Determine (a) the velocity of an electron that has E_k = 6.11 × 10⁻²¹ J, (b) the velocity of a neutron that has E_k = 8.03 × 10⁻²² J, (c) the mass and identity of a subatomic particle moving at 1.447 × 10³ m/s that has E_k = 9.5367 × 10⁻²⁵ J.

Interpretation Introduction

Interpretation:

The velocity of the given atoms, the mass and identity of a subatomic particle should be calculated in the given statement by using the equation of kinetic energy

Concept Introduction:

Energy is the capacity to do work or transfer heat where work is the movement of a body using some force.  The SI unit of energy is joule ($\text{J}$).  Energy is in the form of kinetic energy or potential energy.  Kinetic energy is the energy associated with motion.  Kinetic energy (in joule) is calculated using the formula:

${\text{E}}_{\text{k}}=\frac{\text{1}}{\text{2}}{\text{mu}}^{\text{2}}$

where $\text{m}$ ‒ mass in kilograms; $\text{u}$ – velocity in meters per second.

Answer to Problem 3.8QP

The velocity of an electron that has ${\text{E}}_{\text{k}}\text{= 6}{\text{.11 × 10}}^{-\text{21}}\text{ J}$ is $\text{1}{\text{.16 × 10}}^{\text{5}}\text{ m/s}$.

Explanation of Solution

To find: Determine the velocity of an electron that has ${\text{E}}_{\text{k}}\text{= 6}{\text{.11 × 10}}^{-\text{21}}\text{ J}$ (a)

Kinetic energy (in joule) is calculated using the formula: ${\text{E}}_{\text{k}}=\frac{\text{1}}{\text{2}}{\text{mu}}^{\text{2}}$

where $\text{m}$ ‒ mass in kilograms; $\text{u}$ – velocity in meters per second.  From this equation, velocity in meters per second is calculated using the formula:

$\text{u }=\sqrt{\frac{{\text{2E}}_{\text{k}}}{\text{m}}}$

By considering the given problem, the mass of an electron $\text{m }=\text{ 9}{\text{.10938×10}}^{-\text{28}}\text{ g}$; ${\text{E}}_{\text{k}}\text{= 6}{\text{.11 × 10}}^{-\text{21}}\text{ J}$.

The mass of an electron in kilograms is

$\begin{array}{l}\text{m }=\text{ 9}{\text{.10938 × 10}}^{-\text{28}}\text{ g }\times \frac{\text{1 kg}}{{\text{1 × 10}}^{\text{3}}\text{ g}}\\ \text{m }=\text{ 9}{\text{.10938 × 10}}^{-\text{31}}\text{ kg}\end{array}$

${\text{E}}_{\text{k}}$ value in $\text{6}{\text{.11 × 10}}^{-\text{21}}\text{ J}$ is equal to ${\text{E}}_{\text{k}}$ value in $\text{6}{\text{.11 × 10}}^{-\text{21}}\text{ kg}\cdot {\text{m}}^{\text{2}}{\text{/s}}^{\text{2}}$ which is used for the purpose of making the unit cancellation. Substitute the given values in the formula,

$\begin{array}{l}\text{u }=\sqrt{\frac{\text{2 × (6}{\text{.11 × 10}}^{-\text{21}}\text{ kg}\cdot {\text{m}}^{\text{2}}{\text{/s}}^{\text{2}}\text{)}}{\text{9}{\text{.10938 × 10}}^{-\text{31}}\text{ kg}}}\\ \text{u }=\text{ 1}{\text{.16 × 10}}^{\text{5}}\text{ m/s}\end{array}$

Therefore, the velocity of an electron that has ${\text{E}}_{\text{k}}\text{= 6}{\text{.11 × 10}}^{-\text{21}}\text{ J}$ is $\text{1}{\text{.16 × 10}}^{\text{5}}\text{ m/s}$

Interpretation Introduction

Interpretation:

The velocity of the given atoms, the mass and identity of a subatomic particle should be calculated in the given statement by using the equation of kinetic energy

Concept Introduction:

Energy is the capacity to do work or transfer heat where work is the movement of a body using some force.  The SI unit of energy is joule ($\text{J}$).  Energy is in the form of kinetic energy or potential energy.  Kinetic energy is the energy associated with motion.  Kinetic energy (in joule) is calculated using the formula:

${\text{E}}_{\text{k}}=\frac{\text{1}}{\text{2}}{\text{mu}}^{\text{2}}$

where $\text{m}$ ‒ mass in kilograms; $\text{u}$ – velocity in meters per second.

Answer to Problem 3.8QP

The velocity of a neutron that has ${\text{E}}_{\text{k}}=\text{ 8}{\text{.03 × 10}}^{-\text{22}}\text{ J}$ is $\text{979 m/s}$

Explanation of Solution

To find: Determine the velocity of a neutron that has ${\text{E}}_{\text{k}}=\text{ 8}{\text{.03 × 10}}^{-\text{22}}\text{ J}$ (b)

Kinetic energy (in joule) is calculated using the formula: ${\text{E}}_{\text{k}}=\frac{\text{1}}{\text{2}}{\text{mu}}^{\text{2}}$

Where, $\text{m}$ ‒ mass in kilograms; $\text{u}$ – velocity in meters per second.  From this equation, velocity in meters per second is calculated using the formulae:

$\text{u }=\sqrt{\frac{{\text{2E}}_{\text{k}}}{\text{m}}}$

By considering the given problem, the mass of a neutron $\text{m }=\text{ 1}{\text{.67493×10}}^{-\text{24}}\text{ g}$; ${\text{E}}_{\text{k}}=\text{ 8}{\text{.03 × 10}}^{-\text{22}}\text{ J}$.

The mass of a neutron in kilograms is

$\begin{array}{l}\text{m }=\text{ 1}{\text{.67493 × 10}}^{-\text{24}}\text{ g }\times \frac{\text{1 kg}}{{\text{1 × 10}}^{\text{3}}\text{ g}}\\ \text{m }=\text{ 1}{\text{.67493 × 10}}^{-\text{27}}\text{ kg}\end{array}$

${\text{E}}_{\text{k}}$ value in $\text{8}{\text{.03 × 10}}^{-\text{22}}\text{ J}$ is equal to ${\text{E}}_{\text{k}}$ value in $\text{8}{\text{.03 × 10}}^{-\text{22}}\text{ kg}\cdot {\text{m}}^{\text{2}}{\text{/s}}^{\text{2}}$ which is used for the purpose of making the unit cancellation. Substitute the given values in the formula,

$\begin{array}{l}\text{u }=\sqrt{\frac{\text{2 × (8}{\text{.03 × 10}}^{-\text{22}}\text{ kg}\cdot {\text{m}}^{\text{2}}{\text{/s}}^{\text{2}}\text{)}}{\text{1}{\text{.67493 × 10}}^{-\text{27}}\text{ kg}}}\\ \text{u }=\text{ 979 m/s}\end{array}$

Therefore, the velocity of a neutron that has ${\text{E}}_{\text{k}}=\text{ 8}{\text{.03 × 10}}^{-\text{22}}\text{ J}$ is $\text{979 m/s}$

Interpretation Introduction

Interpretation:

The velocity of the given atoms, the mass and identity of a subatomic particle should be calculated in the given statement by using the equation of kinetic energy

Concept Introduction:

Energy is the capacity to do work or transfer heat where work is the movement of a body using some force.  The SI unit of energy is joule ($\text{J}$).  Energy is in the form of kinetic energy or potential energy.  Kinetic energy is the energy associated with motion.  Kinetic energy (in joule) is calculated using the formula:

${\text{E}}_{\text{k}}=\frac{\text{1}}{\text{2}}{\text{mu}}^{\text{2}}$

where $\text{m}$ ‒ mass in kilograms; $\text{u}$ – velocity in meters per second.

Answer to Problem 3.8QP

The mass and identity of a subatomic particle moving at $\text{1}{\text{.447 × 10}}^{\text{3}}\text{ m/s}$ that has ${\text{E}}_{\text{k}}\text{= 9}{\text{.5367 × 10}}^{-\text{25}}\text{ J}$ are $\text{9}{\text{.109 × 10}}^{-\text{31}}\text{ kg}$ and an electron

Explanation of Solution

To find: Determine the mass and identity of a subatomic particle moving at $\text{1}{\text{.447 × 10}}^{\text{3}}\text{ m/s}$ that has ${\text{E}}_{\text{k}}\text{= 9}{\text{.5367 × 10}}^{-\text{25}}\text{ J}$ (c)

Kinetic energy (in joule) is calculated using the formula: ${\text{E}}_{\text{k}}=\frac{\text{1}}{\text{2}}{\text{mu}}^{\text{2}}$

where $\text{m}$ ‒ mass in kilograms; $\text{u}$ – velocity in meters per second.  From this equation, mass in kilograms is calculated using the formula:

$\text{m }=\frac{{\text{2E}}_{\text{k}}}{{\text{u}}^{\text{2}}}$

By considering the given problem, the mass of a $\text{Kr}$ atom $\text{m }=\text{ 83}\text{.80 amu}$; ${\text{E}}_{\text{k}}\text{= 9}{\text{.5367 × 10}}^{-\text{25}}\text{ J}$.  ${\text{E}}_{\text{k}}$ value in $\text{9}{\text{.5367 × 10}}^{-\text{25}}\text{ J}$ is equal to ${\text{E}}_{\text{k}}$ value in $\text{9}{\text{.5367 × 10}}^{-\text{25}}\text{ kg}\cdot {\text{m}}^{\text{2}}{\text{/s}}^{\text{2}}$ which is used for the purpose of making the unit cancellation.

The mass of a subatomic particle in kilograms is

$\begin{array}{l}\text{m }=\frac{\text{2 × (9}{\text{.5367 × 10}}^{-\text{25}}\text{ kg}\cdot {\text{m}}^{\text{2}}{\text{/s}}^{\text{2}}\text{)}}{{\text{(1}{\text{.477 × 10}}^{\text{3}}\text{ m/s)}}^{\text{2}}}\\ \text{m }=\text{ 9}{\text{.109 × 10}}^{-\text{31}}\text{ kg}\end{array}$

If the mass in $\text{kg}$ is converted into the mass in $\text{g}$, the identity of a subatomic particle will be determined.  The factor for conversion of $\text{kg }\to \text{ g}$ is given as follows:

$\frac{{\text{1 × 10}}^{\text{3}}\text{ g}}{\text{1 kg}}$

By substituting the mass value in the above expression, the identity of a subatomic particle will be determined as follows:

$\begin{array}{l}\text{9}{\text{.109 × 10}}^{-\text{31}}\text{ kg }\times \frac{{\text{1 × 10}}^{\text{3}}\text{ g}}{\text{1 kg}}\\ \text{9}{\text{.109 × 10}}^{-\text{28}}\text{ g}\end{array}$

The subatomic particle with a mass of $\text{9}{\text{.109 × 10}}^{-28}\text{ g}$ is an electron.  Therefore, the mass and identity of a subatomic particle moving at $\text{1}{\text{.447 × 10}}^{\text{3}}\text{ m/s}$ that has ${\text{E}}_{\text{k}}\text{= 9}{\text{.5367 × 10}}^{-\text{25}}\text{ J}$ are $\text{9}{\text{.109 × 10}}^{-\text{31}}\text{ kg}$ and an electron.