Chapter 40, Problem 1OQ

To determine

Rank the wavelengths of the given quantum particles from the largest to smallest.

Answer to Problem 1OQ

The ranking of the wavelengths of the quantum particles is $\underset{\_}{\text{d}>\text{a}=\text{e}>\text{b}>\text{c}}$.

Explanation of Solution

The given particles are,

1. (a) A photon with kinetic energy of $3\text{eV}$.
2. (b) An electron with kinetic energy $3\text{eV}$.
3. (c) A proton with kinetic energy $3\text{eV}$.
4. (d) A photon with kinetic energy of $0.3\text{eV}$.
5. (e) An electron with momentum $\frac{3\text{eV}}{c}$.

Write the expression for the wavelength of the quantum particle.

$\lambda =\frac{h}{p}$ (I)

Here, $\lambda$ is the wavelength of the quantum particle, $h$ is the Planck’s constant, and $p$ is the momentum of the particle.

Write the expression for the momentum of photons.

$p=\frac{E}{c}$ (II)

Here, $E$ is the energy of photon, and $c$ is the speed of light.

From equation (II), photon with higher energy will have higher momentum. Energy of the photon in option (a) is higher than the energy of photon of option (d). From equation (I) it is clear that photon with least energy has longer wavelength. Thus $d>a$.

Write the expression for the relativistic energy of particle.

$E^2=\sqrt{p^2{c}^2+m^2{c}^4}$ (III)

Rewrite expression (III) for $pc$.

$pc={\left(E^2-m^2{c}^4\right)}^{\frac{1}{2}}$ (IV)

Write the expression for kinetic energy of the particle.

$K=E-m{c}^2$ (V)

Solve expression (V) for $E$.

$E=K+m{c}^2$ (VI)

Use expression (VI) in (IV).

$\begin{array}{c}pc={\left({\left[K+m{c}^2\right]}^2-m^2{c}^4\right)}^{\frac{1}{2}}\\ ={\left(K^2+2Km{c}^2\right)}^{\frac{1}{2}}\end{array}$ (VII)

The kinetic energy of both electron and proton of options (b) and (c) are equal. Then from equation (VII), it is clear that momentum of particle is directly proportional to the mass of the particle. Mass of proton is greater than mass of electron. So the momentum of proton is greater than momentum of electron. Again from equation (I), particle with small momentum have longer wavelength. Thus $b>c$.

Substitute $\frac{3\text{eV}}{c}$ for $p$ in equation (II).

$\begin{array}{c}\frac{3\text{eV}}{c}=\frac{E}{c}\\ E=3\text{eV}\end{array}$

Thus the energy of electron of option (e) has energy $3\text{eV}$ which is equal to the energy of particles of option (a). Thus the wavelength of particle of option (e) is equal to the wavelength of option (a). Thus $a=e$.

Thus combining all the results the ranking of the particles based on the wavelength is $\underset{\_}{\text{d}>\text{a}=\text{e}>\text{b}>\text{c}}$.

Conclusion:

Therefore, the ranking of the wavelengths of the quantum particles is $\underset{\_}{\text{d}>\text{a}=\text{e}>\text{b}>\text{c}}$.