Chapter 7, Problem 7.71QP

(a)

Interpretation Introduction

Interpretation: The wavelength of the given objects is to be calculated.

Concept introduction: Wavelength is defined as the equal distance between the two successive troughs and wave crests. It is denoted by a symbol lambda $\left(\lambda \right)$ and is calculated in the unit of meter or nanometer. The wavelength of photons emitted is inversely proportional to the frequency of photon.

To determine: The wavelength of the given object.

Answer to Problem 7.71QP

Solution

The wavelength of a muon is $\underset{\_}{10.82nm}$.

Explanation of Solution

Explanation

Given

Mass of muon (a subatomic particle) is $1.884\times {10}^{-25}\text{g}$.

The conversion of g to kg is done as,

$1\text{g}={10}^{-3}\text{kg}$

Therefore the conversion of $1.884\times {10}^{-25}\text{g}$ to kg is done as,

$\begin{array}{c}1.884\times {10}^{-25}\text{g}=1.884\times {10}^{-25}\times {10}^{-3}\text{kg}\\ =1.884\times {10}^{-28}\text{kg}\end{array}$

The velocity of particle is $325\text{m/s}$.

The wavelength of a moving particle is given by the equation which is represented as,

$\lambda =\frac{h}{mu}$

Where,

• $h$ is the Planck’s constant $\left(6.626\times {10}^{-34}\text{J}\text{.s}\right)$.
• $u$ is the velocity in $\text{m/s}$.
• $m$ is the mass of the particle in $\text{kg}$.
• $\lambda$ is the wavelength of a particle in nm.

Also, $1\text{J}=1\text{kg}{\text{.m}}^{\text{2}}{\text{/s}}^{\text{2}}$

Substitute the values of $h$, $u$ and $m$ in the above equation to calculate the wavelength.

$\begin{array}{c}\lambda =\frac{6.626\times {10}^{-34}\text{J}\text{.s}}{\left(1.884\times {10}^{-28}\text{kg}\right)\left(32\text{5}\text{m/s}\right)}\\ =\frac{6.626\times {10}^{-34}\text{kg}{\text{.m}}^{\text{2}}{\text{/s}}^2\text{.s}}{6.123\times {10}^{-26}\text{kg}\text{.m/s}}\\ =1.082\times {10}^{-8}\text{m}\end{array}$

The wavelength of an object is $1.082\times {10}^{-8}\text{m}$.

The conversion of m to nm is done as,

$1\text{m}={10}^9\text{nm}$

Therefore the conversion of $1.082\times {10}^{-8}\text{m}$ to nm is done as,

$\begin{array}{c}1.082\times {10}^{-8}\text{m}=1.082\times {10}^{-8}\times {10}^9\text{nm}\\ =\underset{\_}{10.82nm}\end{array}$

Hence, the wavelength of a muon is $\underset{\_}{10.82nm}$.

(b)

Interpretation Introduction

To determine: The wavelength of the given object.

Answer to Problem 7.71QP

Solution

The wavelength of an electron is $\underset{\_}{0.1796nm}$.

Explanation of Solution

Explanation

Given

Mass of electron is $9.10938\times {10}^{-28}\text{g}$.

The conversion of g to kg is done as,

$1\text{g}={10}^{-3}\text{kg}$

Therefore the conversion of $9.10938\times {10}^{-28}\text{g}$ to kg is done as,

$\begin{array}{c}9.10938\times {10}^{-28}\text{g}=9.10938\times {10}^{-28}\times {10}^{-3}\text{kg}\\ =9.10938\times {10}^{-31}\text{kg}\end{array}$

The velocity of electron is $4.05\times {10}^6\text{m/s}$.

The wavelength of a moving electron is given by the equation which is represented as,

$\lambda =\frac{h}{mu}$

Where,

• $h$ is the Planck’s constant $\left(6.626\times {10}^{-34}\text{J}\text{.s}\right)$.
• $u$ is the velocity in $\text{m/s}$.
• $m$ is the mass of an electron in $\text{kg}$.
• $\lambda$ is the wavelength of an electron in nm.

Also, $1\text{J}=1\text{kg}{\text{.m}}^{\text{2}}{\text{/s}}^{\text{2}}$

Substitute the values of $h$, $u$ and $m$ in the above equation to calculate the wavelength.

$\begin{array}{c}\lambda =\frac{6.626\times {10}^{-34}\text{J}\text{.s}}{\left(9.10939\times {10}^{-31}\text{kg}\right)\left(4.05\times {10}^6\text{m/s}\right)}\\ =\frac{6.626\times {10}^{-34}\text{kg}{\text{.m}}^{\text{2}}{\text{/s}}^2\text{.s}}{3.689\times {10}^{-24}\text{kg}\text{.m/s}}\\ =1.796\times {10}^{-10}\text{m}\end{array}$

The wavelength of an electron is $1.796\times {10}^{-10}\text{m}$.

The conversion of m to nm is done as,

$1\text{m}={10}^9\text{nm}$

Therefore the conversion of $1.796\times {10}^{-10}\text{m}$ to nm is done as,

$\begin{array}{c}1.796\times {10}^{-10}\text{m}=1.796\times {10}^{-10}\times {10}^9\text{nm}\\ =\underset{\_}{0.1796nm}\end{array}$

Hence, the wavelength of an electron is $\underset{\_}{0.1796nm}$.

(c)

Interpretation Introduction

To determine: The wavelength of the given object.

Answer to Problem 7.71QP

Solution

The wavelength of an athlete is $\underset{\_}{1.235\times 10^{-27}nm}$.

Explanation of Solution

Explanation

Given

Mass of athlete is $80\text{kg}$.

A mile distance is running by athlete in 4 minutes.

Distance is 1 mile.

Time is 4 minutes.

The conversion of mile to m is done as,

$1\text{mile}=1609.34\text{m}$

Therefore conversion of 1 mile to m is done as,

$1\text{mile}=1609.34\text{m}$

The conversion of min to sec is done as,

$1\min =60\sec$

Therefore, the conversion of 4 min to sec is done as,

$\begin{array}{c}4\min =4\times 60\sec \\ =240\sec \end{array}$

The speed of athlete is calculated by using the formula,

$\text{Speed}=\frac{\text{Distance}}{\text{Time}}$

Substitute the value of distance and time in the above formula to calculate the speed of an athlete.

$\begin{array}{c}\text{Speed}=\frac{1609.34\text{m}}{240\mathrm{s}}\\ =6.705\text{m/s}\end{array}$

The speed of an athlete is $6.705\text{m/s}$.

The wavelength is given by the equation which is represented as,

$\lambda =\frac{h}{mu}$

Where,

• $h$ is the Planck’s constant $\left(6.626\times {10}^{-34}\text{J}\text{.s}\right)$.
• $u$ is the velocity in $\text{m/s}$.
• $m$ is the mass in $\text{kg}$.
• $\lambda$ is the wavelength in nm.

Also, $1\text{J}=1\text{kg}{\text{.m}}^{\text{2}}{\text{/s}}^{\text{2}}$

Substitute the values of $h$, $u$ and $m$ in the above equation to calculate the wavelength.

$\begin{array}{c}\lambda =\frac{6.626\times {10}^{-34}\text{J}\text{.s}}{\left(80\text{kg}\right)\left(6.705\text{m/s}\right)}\\ =\frac{6.626\times {10}^{-34}\text{kg}{\text{.m}}^{\text{2}}{\text{/s}}^2\text{.s}}{536.44\text{kg}\text{.m/s}}\\ =1.235\times {10}^{-36}\text{m}\end{array}$

The wavelength of an athlete is $1.235\times {10}^{-36}\text{m}$.

The conversion of m to nm is done as,

$1\text{m}={10}^9\text{nm}$

Therefore the conversion of $1.235\times {10}^{-36}\text{m}$ to nm is done as,

$\begin{array}{c}1.235\times {10}^{-36}\text{m}=1.235\times {10}^{-36}\times {10}^9\text{nm}\\ =\underset{\_}{1.235\times 10^{-27}nm}\end{array}$

Hence, the wavelength of an athlete is $\underset{\_}{1.235\times 10^{-27}nm}$.

(d)

Interpretation Introduction

To determine: The wavelength of the given object.

Answer to Problem 7.71QP

Solution

The wavelength of an object is $\underset{\_}{3.681\times 10^{-54}nm}$.

Explanation of Solution

Explanation

Given

The mass of earth is $6.0\times {10}^{27}\text{g}$.

The conversion of g to kg is done as,

$1\text{g}={10}^{-3}\text{kg}$

Therefore the conversion of $6.0\times {10}^{27}\text{g}$ to kg is done as,

$\begin{array}{c}6.0\times {10}^{27}\text{g}=6.0\times {10}^{27}\times {10}^{-3}\text{kg}\\ =6.0\times {10}^{24}\text{kg}\end{array}$

The velocity is $3.0\times {10}^4\text{m/s}$.

The wavelength is given by the equation which is represented as,

$\lambda =\frac{h}{mu}$

Where,

• $h$ is the Planck’s constant $\left(6.626\times {10}^{-34}\text{J}\text{.s}\right)$.
• $u$ is the velocity in $\text{m/s}$.
• $m$ is the mass in $\text{kg}$.
• $\lambda$ is the wavelength in nm.

Also, $1\text{J}=1\text{kg}{\text{.m}}^{\text{2}}{\text{/s}}^{\text{2}}$

Substitute the values of $h$, $u$ and $m$ in the above equation to calculate the wavelength.

$\begin{array}{c}\lambda =\frac{6.626\times {10}^{-34}\text{J}\text{.s}}{\left(6.0\times {10}^{24}\text{kg}\right)\left(3.0\times {10}^4\text{m/s}\right)}\\ =\frac{6.626\times {10}^{-34}\text{kg}{\text{.m}}^{\text{2}}{\text{/s}}^2\text{.s}}{18\times {10}^{28}\text{kg}\text{.m/s}}\\ =3.681\times {10}^{-63}\text{m}\end{array}$

The wavelength is $3.681\times {10}^{-63}\text{m}$

The conversion of m to nm is done as,

$1\text{m}={10}^9\text{nm}$

Therefore the conversion of $3.681\times {10}^{-63}\text{m}$ to nm is done as,

$\begin{array}{c}3.681\times {10}^{-63}\text{m}=3.681\times {10}^{-63}\times {10}^9\text{nm}\\ =\underset{\_}{3.681\times 10^{-54}nm}\end{array}$

Hence, the wavelength is $\underset{\_}{3.681\times 10^{-54}nm}$.

Conclusion

1. a. The wavelength of a muon is $\underset{\_}{10.82nm}$.
2. b. The wavelength of an electron is $\underset{\_}{0.1796nm}$.
3. c. The wavelength of an athlete is $\underset{\_}{1.235\times 10^{-27}nm}$.
4. d. The wavelength of an object is $\underset{\_}{3.681\times 10^{-54}nm}$.