To show that the frequency of the electron’s orbital motion is f e = m e k 2 Z 2 e 4 2 π ℏ 3 ( 1 n 3 ) .

Question

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Answer 1

Question

Chapter 4, Problem 31P

(a)

To determine

To show that the frequency of the electron’s orbital motion is $fe=mek2Z2e42πℏ3(1n3)$ .

(a)

Expert Solution

Answer to Problem 31P

It is showed that the frequency of the electron’s orbital motion is $fe=mek2Z2e42πℏ3(1n3)$ .

Explanation of Solution

Write the expression for the conservation of angular momentum of the electron.

$mevr=nℏ$

Here, $me$ is the mass of the electron, $v$ is the orbital speed of the electron, $r$ is the radius of the orbit, $n$ is quantum number of the orbit and $ℏ$ is the reduced Planck’s constant.

Rewrite the above equation for $v$ .

$v=nℏmer$ (I)

Write the equation for the frequency of the electron.

$fe=v2πr$

Here, $fe$ is the frequency of the electron’s orbital motion.

Put equation (I) in the above equation.

$fe=nℏmer2πr=nℏ2πmer2$ (II)

Write the equation for the radius of an orbit.

$r=a0n2$

Here, $a0$ is the Bohr radius.

Put the above equation in equation (II).

$fe=nℏ2πme(a0n2)2=ℏ2πmea02n3$ (III)

Write the equation for the Bohr radius.

$a0=ℏ2meZke2$ (IV)

Here, $Z$ is the atomic number, $k$ is the Coulomb constant and $e$ is the magnitude of the charge of the electron.

Put the above equation in equation (III).

$fe=ℏ2πme(ℏ2meZke2)2n3=ℏme2Z2k2e42πmeℏ4n3=mek2Z2e42πℏ3(1n3)$

Conclusion:

Therefore, it is showed that the frequency of the electron’s orbital motion is $fe=mek2Z2e42πℏ3(1n3)$

(b)

To determine

To show that frequency of the photon emitted when an electron jumps from an outer to inner orbit can be written as $fe=mek2Z2e42πℏ3ni+nf2ni2nf2(ni−nf)$ and $1ni3<ni+nf2ni2nf2<1nf3$ for electronic transitions between adjacent orbits.

(b)

Expert Solution

Answer to Problem 31P

It is showed that frequency of the photon emitted when an electron jumps from an outer to inner orbit can be written as $fphoton=mek2Z2e42πℏ3ni+nf2ni2nf2(ni−nf)$ and $1ni3<ni+nf2ni2nf2<1nf3$ for electronic transitions between adjacent orbits.

Explanation of Solution

Write the equation for the energy of the photon emitted.

$ΔE=hfphoton$

Here, $ΔE$ is the energy of the photon emitted, $h$ is the Planck’s constant and $fphoton$ is the frequency of the photon.

Write the equation for the energy of the photon emitted when an electron jumps from an outer to inner orbit.

$ΔE=h(kZ2e22a0h)(1nf2−1ni2)$

Here, $ni$ is the quantum number of the initial level and $nf$ is the quantum number of the final level.

Compare the above two equations to write the expression for $fphoton$ .

$fphoton=(kZ2e22a0h)(1nf2−1ni2)=(kZ2e22a0h)(ni2−nf2ni2nf2)=(kZ2e22a0h)(ni+nf)(ni−nf)ni2nf2$

Put the equation for Bohr radius in the above equation.

$fphoton=(kZ2e22(ℏ2meke2)h)(ni+nf)(ni−nf)ni2nf2=mek2Z2e42ℏ2h(ni+nf)(ni−nf)ni2nf2=mek2Z2e42ℏ2(h2π)(2π)(ni+nf)(ni−nf)ni2nf2=mek2Z2e42πℏ3(ni+nf)(ni−nf)2ni2nf2$ (V)

Write the relationship between $ni$ and $nf$ for transitions between adjacent orbits.

$ni−nf=1$

Put the above equation in equation (V).

$fphoton=mek2Z2e42πℏ3(ni+nf)(1)2ni2nf2=mek2Z2e42πℏ3(ni+nf)2ni2nf2$ (VI)

Conclusion:

For $ni=2,nf=1,ni=3,nf=2, etc.,$ using equation (VI) gives that $1ni3<ni+nf2ni2nf2<1nf3$ .

Therefore, it is showed that frequency of the photon emitted when an electron jumps from an outer to inner orbit can be written as $fphoton=mek2Z2e42πℏ3ni+nf2ni2nf2(ni−nf)$ and $1ni3<ni+nf2ni2nf2<1nf3$ for electronic transitions between adjacent orbits.

(c)

To determine

The conclusion regarding the frequency of the emitted radiation compared with the frequencies of orbital revolution in the initial and final states and the result as $ni→∞$ .

(c)

Expert Solution

Answer to Problem 31P

Frequency of the emitted radiation is in between the initial and the final orbital frequency and the frequency of the emitted radiation becomes the orbital frequency as $ni→∞$ .

Explanation of Solution

The expression for the orbital frequency of the electron is found in part (a). The equation for the frequency of the emitted photon is found in part (b). Comparison of the two equations shows that the frequency of the emitted radiation is in between the initial orbital frequency and the final orbital frequency.

As $ni→∞$ , the initial orbital frequency and the final orbital frequency squeeze closer together. This will make the frequency of the emitted radiation equal to the orbital frequency. This result is in agreement with Bohr’s correspondence principle.

Conclusion:

Thus, the frequency of the emitted radiation is in between the initial and the final orbital frequency and the frequency of the emitted radiation becomes the orbital frequency as $ni→∞$ .

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Answer 2

Question

Chapter 4, Problem 31P

(a)

To determine

To show that the frequency of the electron’s orbital motion is $fe=mek2Z2e42πℏ3(1n3)$ .

(a)

Expert Solution

Answer to Problem 31P

It is showed that the frequency of the electron’s orbital motion is $fe=mek2Z2e42πℏ3(1n3)$ .

Explanation of Solution

Write the expression for the conservation of angular momentum of the electron.

$mevr=nℏ$

Here, $me$ is the mass of the electron, $v$ is the orbital speed of the electron, $r$ is the radius of the orbit, $n$ is quantum number of the orbit and $ℏ$ is the reduced Planck’s constant.

Rewrite the above equation for $v$ .

$v=nℏmer$ (I)

Write the equation for the frequency of the electron.

$fe=v2πr$

Here, $fe$ is the frequency of the electron’s orbital motion.

Put equation (I) in the above equation.

$fe=nℏmer2πr=nℏ2πmer2$ (II)

Write the equation for the radius of an orbit.

$r=a0n2$

Here, $a0$ is the Bohr radius.

Put the above equation in equation (II).

$fe=nℏ2πme(a0n2)2=ℏ2πmea02n3$ (III)

Write the equation for the Bohr radius.

$a0=ℏ2meZke2$ (IV)

Here, $Z$ is the atomic number, $k$ is the Coulomb constant and $e$ is the magnitude of the charge of the electron.

Put the above equation in equation (III).

$fe=ℏ2πme(ℏ2meZke2)2n3=ℏme2Z2k2e42πmeℏ4n3=mek2Z2e42πℏ3(1n3)$

Conclusion:

Therefore, it is showed that the frequency of the electron’s orbital motion is $fe=mek2Z2e42πℏ3(1n3)$

(b)

To determine

To show that frequency of the photon emitted when an electron jumps from an outer to inner orbit can be written as $fe=mek2Z2e42πℏ3ni+nf2ni2nf2(ni−nf)$ and $1ni3<ni+nf2ni2nf2<1nf3$ for electronic transitions between adjacent orbits.

(b)

Expert Solution

Answer to Problem 31P

It is showed that frequency of the photon emitted when an electron jumps from an outer to inner orbit can be written as $fphoton=mek2Z2e42πℏ3ni+nf2ni2nf2(ni−nf)$ and $1ni3<ni+nf2ni2nf2<1nf3$ for electronic transitions between adjacent orbits.

Explanation of Solution

Write the equation for the energy of the photon emitted.

$ΔE=hfphoton$

Here, $ΔE$ is the energy of the photon emitted, $h$ is the Planck’s constant and $fphoton$ is the frequency of the photon.

Write the equation for the energy of the photon emitted when an electron jumps from an outer to inner orbit.

$ΔE=h(kZ2e22a0h)(1nf2−1ni2)$

Here, $ni$ is the quantum number of the initial level and $nf$ is the quantum number of the final level.

Compare the above two equations to write the expression for $fphoton$ .

$fphoton=(kZ2e22a0h)(1nf2−1ni2)=(kZ2e22a0h)(ni2−nf2ni2nf2)=(kZ2e22a0h)(ni+nf)(ni−nf)ni2nf2$

Put the equation for Bohr radius in the above equation.

$fphoton=(kZ2e22(ℏ2meke2)h)(ni+nf)(ni−nf)ni2nf2=mek2Z2e42ℏ2h(ni+nf)(ni−nf)ni2nf2=mek2Z2e42ℏ2(h2π)(2π)(ni+nf)(ni−nf)ni2nf2=mek2Z2e42πℏ3(ni+nf)(ni−nf)2ni2nf2$ (V)

Write the relationship between $ni$ and $nf$ for transitions between adjacent orbits.

$ni−nf=1$

Put the above equation in equation (V).

$fphoton=mek2Z2e42πℏ3(ni+nf)(1)2ni2nf2=mek2Z2e42πℏ3(ni+nf)2ni2nf2$ (VI)

Conclusion:

For $ni=2,nf=1,ni=3,nf=2, etc.,$ using equation (VI) gives that $1ni3<ni+nf2ni2nf2<1nf3$ .

Therefore, it is showed that frequency of the photon emitted when an electron jumps from an outer to inner orbit can be written as $fphoton=mek2Z2e42πℏ3ni+nf2ni2nf2(ni−nf)$ and $1ni3<ni+nf2ni2nf2<1nf3$ for electronic transitions between adjacent orbits.

(c)

To determine

The conclusion regarding the frequency of the emitted radiation compared with the frequencies of orbital revolution in the initial and final states and the result as $ni→∞$ .

(c)

Expert Solution

Answer to Problem 31P

Frequency of the emitted radiation is in between the initial and the final orbital frequency and the frequency of the emitted radiation becomes the orbital frequency as $ni→∞$ .

Explanation of Solution

The expression for the orbital frequency of the electron is found in part (a). The equation for the frequency of the emitted photon is found in part (b). Comparison of the two equations shows that the frequency of the emitted radiation is in between the initial orbital frequency and the final orbital frequency.

As $ni→∞$ , the initial orbital frequency and the final orbital frequency squeeze closer together. This will make the frequency of the emitted radiation equal to the orbital frequency. This result is in agreement with Bohr’s correspondence principle.

Conclusion:

Thus, the frequency of the emitted radiation is in between the initial and the final orbital frequency and the frequency of the emitted radiation becomes the orbital frequency as $ni→∞$ .

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Answer 3

Question

Chapter 4, Problem 31P

(a)

To determine

To show that the frequency of the electron’s orbital motion is $fe=mek2Z2e42πℏ3(1n3)$ .

(a)

Expert Solution

Answer to Problem 31P

It is showed that the frequency of the electron’s orbital motion is $fe=mek2Z2e42πℏ3(1n3)$ .

Explanation of Solution

Write the expression for the conservation of angular momentum of the electron.

$mevr=nℏ$

Here, $me$ is the mass of the electron, $v$ is the orbital speed of the electron, $r$ is the radius of the orbit, $n$ is quantum number of the orbit and $ℏ$ is the reduced Planck’s constant.

Rewrite the above equation for $v$ .

$v=nℏmer$ (I)

Write the equation for the frequency of the electron.

$fe=v2πr$

Here, $fe$ is the frequency of the electron’s orbital motion.

Put equation (I) in the above equation.

$fe=nℏmer2πr=nℏ2πmer2$ (II)

Write the equation for the radius of an orbit.

$r=a0n2$

Here, $a0$ is the Bohr radius.

Put the above equation in equation (II).

$fe=nℏ2πme(a0n2)2=ℏ2πmea02n3$ (III)

Write the equation for the Bohr radius.

$a0=ℏ2meZke2$ (IV)

Here, $Z$ is the atomic number, $k$ is the Coulomb constant and $e$ is the magnitude of the charge of the electron.

Put the above equation in equation (III).

$fe=ℏ2πme(ℏ2meZke2)2n3=ℏme2Z2k2e42πmeℏ4n3=mek2Z2e42πℏ3(1n3)$

Conclusion:

Therefore, it is showed that the frequency of the electron’s orbital motion is $fe=mek2Z2e42πℏ3(1n3)$

(b)

To determine

To show that frequency of the photon emitted when an electron jumps from an outer to inner orbit can be written as $fe=mek2Z2e42πℏ3ni+nf2ni2nf2(ni−nf)$ and $1ni3<ni+nf2ni2nf2<1nf3$ for electronic transitions between adjacent orbits.

(b)

Expert Solution

Answer to Problem 31P

It is showed that frequency of the photon emitted when an electron jumps from an outer to inner orbit can be written as $fphoton=mek2Z2e42πℏ3ni+nf2ni2nf2(ni−nf)$ and $1ni3<ni+nf2ni2nf2<1nf3$ for electronic transitions between adjacent orbits.

Explanation of Solution

Write the equation for the energy of the photon emitted.

$ΔE=hfphoton$

Here, $ΔE$ is the energy of the photon emitted, $h$ is the Planck’s constant and $fphoton$ is the frequency of the photon.

Write the equation for the energy of the photon emitted when an electron jumps from an outer to inner orbit.

$ΔE=h(kZ2e22a0h)(1nf2−1ni2)$

Here, $ni$ is the quantum number of the initial level and $nf$ is the quantum number of the final level.

Compare the above two equations to write the expression for $fphoton$ .

$fphoton=(kZ2e22a0h)(1nf2−1ni2)=(kZ2e22a0h)(ni2−nf2ni2nf2)=(kZ2e22a0h)(ni+nf)(ni−nf)ni2nf2$

Put the equation for Bohr radius in the above equation.

$fphoton=(kZ2e22(ℏ2meke2)h)(ni+nf)(ni−nf)ni2nf2=mek2Z2e42ℏ2h(ni+nf)(ni−nf)ni2nf2=mek2Z2e42ℏ2(h2π)(2π)(ni+nf)(ni−nf)ni2nf2=mek2Z2e42πℏ3(ni+nf)(ni−nf)2ni2nf2$ (V)

Write the relationship between $ni$ and $nf$ for transitions between adjacent orbits.

$ni−nf=1$

Put the above equation in equation (V).

$fphoton=mek2Z2e42πℏ3(ni+nf)(1)2ni2nf2=mek2Z2e42πℏ3(ni+nf)2ni2nf2$ (VI)

Conclusion:

For $ni=2,nf=1,ni=3,nf=2, etc.,$ using equation (VI) gives that $1ni3<ni+nf2ni2nf2<1nf3$ .

Therefore, it is showed that frequency of the photon emitted when an electron jumps from an outer to inner orbit can be written as $fphoton=mek2Z2e42πℏ3ni+nf2ni2nf2(ni−nf)$ and $1ni3<ni+nf2ni2nf2<1nf3$ for electronic transitions between adjacent orbits.

(c)

To determine

The conclusion regarding the frequency of the emitted radiation compared with the frequencies of orbital revolution in the initial and final states and the result as $ni→∞$ .

(c)

Expert Solution

Answer to Problem 31P

Frequency of the emitted radiation is in between the initial and the final orbital frequency and the frequency of the emitted radiation becomes the orbital frequency as $ni→∞$ .

Explanation of Solution

The expression for the orbital frequency of the electron is found in part (a). The equation for the frequency of the emitted photon is found in part (b). Comparison of the two equations shows that the frequency of the emitted radiation is in between the initial orbital frequency and the final orbital frequency.

As $ni→∞$ , the initial orbital frequency and the final orbital frequency squeeze closer together. This will make the frequency of the emitted radiation equal to the orbital frequency. This result is in agreement with Bohr’s correspondence principle.

Conclusion:

Thus, the frequency of the emitted radiation is in between the initial and the final orbital frequency and the frequency of the emitted radiation becomes the orbital frequency as $ni→∞$ .

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Answer 4

Question

Chapter 4, Problem 31P

(a)

To determine

To show that the frequency of the electron’s orbital motion is $fe=mek2Z2e42πℏ3(1n3)$ .

(a)

Expert Solution

Answer to Problem 31P

It is showed that the frequency of the electron’s orbital motion is $fe=mek2Z2e42πℏ3(1n3)$ .

Explanation of Solution

Write the expression for the conservation of angular momentum of the electron.

$mevr=nℏ$

Here, $me$ is the mass of the electron, $v$ is the orbital speed of the electron, $r$ is the radius of the orbit, $n$ is quantum number of the orbit and $ℏ$ is the reduced Planck’s constant.

Rewrite the above equation for $v$ .

$v=nℏmer$ (I)

Write the equation for the frequency of the electron.

$fe=v2πr$

Here, $fe$ is the frequency of the electron’s orbital motion.

Put equation (I) in the above equation.

$fe=nℏmer2πr=nℏ2πmer2$ (II)

Write the equation for the radius of an orbit.

$r=a0n2$

Here, $a0$ is the Bohr radius.

Put the above equation in equation (II).

$fe=nℏ2πme(a0n2)2=ℏ2πmea02n3$ (III)

Write the equation for the Bohr radius.

$a0=ℏ2meZke2$ (IV)

Here, $Z$ is the atomic number, $k$ is the Coulomb constant and $e$ is the magnitude of the charge of the electron.

Put the above equation in equation (III).

$fe=ℏ2πme(ℏ2meZke2)2n3=ℏme2Z2k2e42πmeℏ4n3=mek2Z2e42πℏ3(1n3)$

Conclusion:

Therefore, it is showed that the frequency of the electron’s orbital motion is $fe=mek2Z2e42πℏ3(1n3)$

(b)

To determine

To show that frequency of the photon emitted when an electron jumps from an outer to inner orbit can be written as $fe=mek2Z2e42πℏ3ni+nf2ni2nf2(ni−nf)$ and $1ni3<ni+nf2ni2nf2<1nf3$ for electronic transitions between adjacent orbits.

(b)

Expert Solution

Answer to Problem 31P

It is showed that frequency of the photon emitted when an electron jumps from an outer to inner orbit can be written as $fphoton=mek2Z2e42πℏ3ni+nf2ni2nf2(ni−nf)$ and $1ni3<ni+nf2ni2nf2<1nf3$ for electronic transitions between adjacent orbits.

Explanation of Solution

Write the equation for the energy of the photon emitted.

$ΔE=hfphoton$

Here, $ΔE$ is the energy of the photon emitted, $h$ is the Planck’s constant and $fphoton$ is the frequency of the photon.

Write the equation for the energy of the photon emitted when an electron jumps from an outer to inner orbit.

$ΔE=h(kZ2e22a0h)(1nf2−1ni2)$

Here, $ni$ is the quantum number of the initial level and $nf$ is the quantum number of the final level.

Compare the above two equations to write the expression for $fphoton$ .

$fphoton=(kZ2e22a0h)(1nf2−1ni2)=(kZ2e22a0h)(ni2−nf2ni2nf2)=(kZ2e22a0h)(ni+nf)(ni−nf)ni2nf2$

Put the equation for Bohr radius in the above equation.

$fphoton=(kZ2e22(ℏ2meke2)h)(ni+nf)(ni−nf)ni2nf2=mek2Z2e42ℏ2h(ni+nf)(ni−nf)ni2nf2=mek2Z2e42ℏ2(h2π)(2π)(ni+nf)(ni−nf)ni2nf2=mek2Z2e42πℏ3(ni+nf)(ni−nf)2ni2nf2$ (V)

Write the relationship between $ni$ and $nf$ for transitions between adjacent orbits.

$ni−nf=1$

Put the above equation in equation (V).

$fphoton=mek2Z2e42πℏ3(ni+nf)(1)2ni2nf2=mek2Z2e42πℏ3(ni+nf)2ni2nf2$ (VI)

Conclusion:

For $ni=2,nf=1,ni=3,nf=2, etc.,$ using equation (VI) gives that $1ni3<ni+nf2ni2nf2<1nf3$ .

Therefore, it is showed that frequency of the photon emitted when an electron jumps from an outer to inner orbit can be written as $fphoton=mek2Z2e42πℏ3ni+nf2ni2nf2(ni−nf)$ and $1ni3<ni+nf2ni2nf2<1nf3$ for electronic transitions between adjacent orbits.

(c)

To determine

The conclusion regarding the frequency of the emitted radiation compared with the frequencies of orbital revolution in the initial and final states and the result as $ni→∞$ .

(c)

Expert Solution

Answer to Problem 31P

Frequency of the emitted radiation is in between the initial and the final orbital frequency and the frequency of the emitted radiation becomes the orbital frequency as $ni→∞$ .

Explanation of Solution

The expression for the orbital frequency of the electron is found in part (a). The equation for the frequency of the emitted photon is found in part (b). Comparison of the two equations shows that the frequency of the emitted radiation is in between the initial orbital frequency and the final orbital frequency.

As $ni→∞$ , the initial orbital frequency and the final orbital frequency squeeze closer together. This will make the frequency of the emitted radiation equal to the orbital frequency. This result is in agreement with Bohr’s correspondence principle.

Conclusion:

Thus, the frequency of the emitted radiation is in between the initial and the final orbital frequency and the frequency of the emitted radiation becomes the orbital frequency as $ni→∞$ .

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Answer 5

Chapter 4, Problem 31P

(a)

To determine

To show that the frequency of the electron’s orbital motion is $fe=mek2Z2e42πℏ3(1n3)$ .

(a)

Expert Solution

Answer to Problem 31P

It is showed that the frequency of the electron’s orbital motion is $fe=mek2Z2e42πℏ3(1n3)$ .

Explanation of Solution

Write the expression for the conservation of angular momentum of the electron.

$mevr=nℏ$

Here, $me$ is the mass of the electron, $v$ is the orbital speed of the electron, $r$ is the radius of the orbit, $n$ is quantum number of the orbit and $ℏ$ is the reduced Planck’s constant.

Rewrite the above equation for $v$ .

$v=nℏmer$ (I)

Write the equation for the frequency of the electron.

$fe=v2πr$

Here, $fe$ is the frequency of the electron’s orbital motion.

Put equation (I) in the above equation.

$fe=nℏmer2πr=nℏ2πmer2$ (II)

Write the equation for the radius of an orbit.

$r=a0n2$

Here, $a0$ is the Bohr radius.

Put the above equation in equation (II).

$fe=nℏ2πme(a0n2)2=ℏ2πmea02n3$ (III)

Write the equation for the Bohr radius.

$a0=ℏ2meZke2$ (IV)

Here, $Z$ is the atomic number, $k$ is the Coulomb constant and $e$ is the magnitude of the charge of the electron.

Put the above equation in equation (III).

$fe=ℏ2πme(ℏ2meZke2)2n3=ℏme2Z2k2e42πmeℏ4n3=mek2Z2e42πℏ3(1n3)$

Conclusion:

Therefore, it is showed that the frequency of the electron’s orbital motion is $fe=mek2Z2e42πℏ3(1n3)$

(b)

To determine

To show that frequency of the photon emitted when an electron jumps from an outer to inner orbit can be written as $fe=mek2Z2e42πℏ3ni+nf2ni2nf2(ni−nf)$ and $1ni3<ni+nf2ni2nf2<1nf3$ for electronic transitions between adjacent orbits.

(b)

Expert Solution

Answer to Problem 31P

It is showed that frequency of the photon emitted when an electron jumps from an outer to inner orbit can be written as $fphoton=mek2Z2e42πℏ3ni+nf2ni2nf2(ni−nf)$ and $1ni3<ni+nf2ni2nf2<1nf3$ for electronic transitions between adjacent orbits.

Explanation of Solution

Write the equation for the energy of the photon emitted.

$ΔE=hfphoton$

Here, $ΔE$ is the energy of the photon emitted, $h$ is the Planck’s constant and $fphoton$ is the frequency of the photon.

Write the equation for the energy of the photon emitted when an electron jumps from an outer to inner orbit.

$ΔE=h(kZ2e22a0h)(1nf2−1ni2)$

Here, $ni$ is the quantum number of the initial level and $nf$ is the quantum number of the final level.

Compare the above two equations to write the expression for $fphoton$ .

$fphoton=(kZ2e22a0h)(1nf2−1ni2)=(kZ2e22a0h)(ni2−nf2ni2nf2)=(kZ2e22a0h)(ni+nf)(ni−nf)ni2nf2$

Put the equation for Bohr radius in the above equation.

$fphoton=(kZ2e22(ℏ2meke2)h)(ni+nf)(ni−nf)ni2nf2=mek2Z2e42ℏ2h(ni+nf)(ni−nf)ni2nf2=mek2Z2e42ℏ2(h2π)(2π)(ni+nf)(ni−nf)ni2nf2=mek2Z2e42πℏ3(ni+nf)(ni−nf)2ni2nf2$ (V)

Write the relationship between $ni$ and $nf$ for transitions between adjacent orbits.

$ni−nf=1$

Put the above equation in equation (V).

$fphoton=mek2Z2e42πℏ3(ni+nf)(1)2ni2nf2=mek2Z2e42πℏ3(ni+nf)2ni2nf2$ (VI)

Conclusion:

For $ni=2,nf=1,ni=3,nf=2, etc.,$ using equation (VI) gives that $1ni3<ni+nf2ni2nf2<1nf3$ .

Therefore, it is showed that frequency of the photon emitted when an electron jumps from an outer to inner orbit can be written as $fphoton=mek2Z2e42πℏ3ni+nf2ni2nf2(ni−nf)$ and $1ni3<ni+nf2ni2nf2<1nf3$ for electronic transitions between adjacent orbits.

(c)

To determine

The conclusion regarding the frequency of the emitted radiation compared with the frequencies of orbital revolution in the initial and final states and the result as $ni→∞$ .

(c)

Expert Solution

Answer to Problem 31P

Frequency of the emitted radiation is in between the initial and the final orbital frequency and the frequency of the emitted radiation becomes the orbital frequency as $ni→∞$ .

Explanation of Solution

The expression for the orbital frequency of the electron is found in part (a). The equation for the frequency of the emitted photon is found in part (b). Comparison of the two equations shows that the frequency of the emitted radiation is in between the initial orbital frequency and the final orbital frequency.

As $ni→∞$ , the initial orbital frequency and the final orbital frequency squeeze closer together. This will make the frequency of the emitted radiation equal to the orbital frequency. This result is in agreement with Bohr’s correspondence principle.

Conclusion:

Thus, the frequency of the emitted radiation is in between the initial and the final orbital frequency and the frequency of the emitted radiation becomes the orbital frequency as $ni→∞$ .

Answer 6

Chapter 4, Problem 31P

(a)

To determine

To show that the frequency of the electron’s orbital motion is $fe=mek2Z2e42πℏ3(1n3)$ .

(a)

Expert Solution

Answer to Problem 31P

It is showed that the frequency of the electron’s orbital motion is $fe=mek2Z2e42πℏ3(1n3)$ .

Explanation of Solution

Write the expression for the conservation of angular momentum of the electron.

$mevr=nℏ$

Here, $me$ is the mass of the electron, $v$ is the orbital speed of the electron, $r$ is the radius of the orbit, $n$ is quantum number of the orbit and $ℏ$ is the reduced Planck’s constant.

Rewrite the above equation for $v$ .

$v=nℏmer$ (I)

Write the equation for the frequency of the electron.

$fe=v2πr$

Here, $fe$ is the frequency of the electron’s orbital motion.

Put equation (I) in the above equation.

$fe=nℏmer2πr=nℏ2πmer2$ (II)

Write the equation for the radius of an orbit.

$r=a0n2$

Here, $a0$ is the Bohr radius.

Put the above equation in equation (II).

$fe=nℏ2πme(a0n2)2=ℏ2πmea02n3$ (III)

Write the equation for the Bohr radius.

$a0=ℏ2meZke2$ (IV)

Here, $Z$ is the atomic number, $k$ is the Coulomb constant and $e$ is the magnitude of the charge of the electron.

Put the above equation in equation (III).

$fe=ℏ2πme(ℏ2meZke2)2n3=ℏme2Z2k2e42πmeℏ4n3=mek2Z2e42πℏ3(1n3)$

Conclusion:

Therefore, it is showed that the frequency of the electron’s orbital motion is $fe=mek2Z2e42πℏ3(1n3)$

Answer 7

Chapter 4, Problem 31P

(a)

To determine

To show that the frequency of the electron’s orbital motion is $fe=mek2Z2e42πℏ3(1n3)$ .

Answer 8

(a)

Expert Solution

Answer to Problem 31P

It is showed that the frequency of the electron’s orbital motion is $fe=mek2Z2e42πℏ3(1n3)$ .

Answer 9

(a)

Expert Solution

Answer 10

(a)

Expert Solution

Answer 11

Expert Solution

Answer 12

Explanation of Solution

Write the expression for the conservation of angular momentum of the electron.

$mevr=nℏ$

Here, $me$ is the mass of the electron, $v$ is the orbital speed of the electron, $r$ is the radius of the orbit, $n$ is quantum number of the orbit and $ℏ$ is the reduced Planck’s constant.

Rewrite the above equation for $v$ .

$v=nℏmer$ (I)

Write the equation for the frequency of the electron.

$fe=v2πr$

Here, $fe$ is the frequency of the electron’s orbital motion.

Put equation (I) in the above equation.

$fe=nℏmer2πr=nℏ2πmer2$ (II)

Write the equation for the radius of an orbit.

$r=a0n2$

Here, $a0$ is the Bohr radius.

Put the above equation in equation (II).

$fe=nℏ2πme(a0n2)2=ℏ2πmea02n3$ (III)

Write the equation for the Bohr radius.

$a0=ℏ2meZke2$ (IV)

Here, $Z$ is the atomic number, $k$ is the Coulomb constant and $e$ is the magnitude of the charge of the electron.

Put the above equation in equation (III).

$fe=ℏ2πme(ℏ2meZke2)2n3=ℏme2Z2k2e42πmeℏ4n3=mek2Z2e42πℏ3(1n3)$

Conclusion:

Therefore, it is showed that the frequency of the electron’s orbital motion is $fe=mek2Z2e42πℏ3(1n3)$

Answer 13

(b)

To determine

To show that frequency of the photon emitted when an electron jumps from an outer to inner orbit can be written as $fe=mek2Z2e42πℏ3ni+nf2ni2nf2(ni−nf)$ and $1ni3<ni+nf2ni2nf2<1nf3$ for electronic transitions between adjacent orbits.

(b)

Expert Solution

Answer to Problem 31P

It is showed that frequency of the photon emitted when an electron jumps from an outer to inner orbit can be written as $fphoton=mek2Z2e42πℏ3ni+nf2ni2nf2(ni−nf)$ and $1ni3<ni+nf2ni2nf2<1nf3$ for electronic transitions between adjacent orbits.

Explanation of Solution

Write the equation for the energy of the photon emitted.

$ΔE=hfphoton$

Here, $ΔE$ is the energy of the photon emitted, $h$ is the Planck’s constant and $fphoton$ is the frequency of the photon.

Write the equation for the energy of the photon emitted when an electron jumps from an outer to inner orbit.

$ΔE=h(kZ2e22a0h)(1nf2−1ni2)$

Here, $ni$ is the quantum number of the initial level and $nf$ is the quantum number of the final level.

Compare the above two equations to write the expression for $fphoton$ .

$fphoton=(kZ2e22a0h)(1nf2−1ni2)=(kZ2e22a0h)(ni2−nf2ni2nf2)=(kZ2e22a0h)(ni+nf)(ni−nf)ni2nf2$

Put the equation for Bohr radius in the above equation.

$fphoton=(kZ2e22(ℏ2meke2)h)(ni+nf)(ni−nf)ni2nf2=mek2Z2e42ℏ2h(ni+nf)(ni−nf)ni2nf2=mek2Z2e42ℏ2(h2π)(2π)(ni+nf)(ni−nf)ni2nf2=mek2Z2e42πℏ3(ni+nf)(ni−nf)2ni2nf2$ (V)

Write the relationship between $ni$ and $nf$ for transitions between adjacent orbits.

$ni−nf=1$

Put the above equation in equation (V).

$fphoton=mek2Z2e42πℏ3(ni+nf)(1)2ni2nf2=mek2Z2e42πℏ3(ni+nf)2ni2nf2$ (VI)

Conclusion:

For $ni=2,nf=1,ni=3,nf=2, etc.,$ using equation (VI) gives that $1ni3<ni+nf2ni2nf2<1nf3$ .

Therefore, it is showed that frequency of the photon emitted when an electron jumps from an outer to inner orbit can be written as $fphoton=mek2Z2e42πℏ3ni+nf2ni2nf2(ni−nf)$ and $1ni3<ni+nf2ni2nf2<1nf3$ for electronic transitions between adjacent orbits.

Answer 14

(b)

To determine

To show that frequency of the photon emitted when an electron jumps from an outer to inner orbit can be written as $fe=mek2Z2e42πℏ3ni+nf2ni2nf2(ni−nf)$ and $1ni3<ni+nf2ni2nf2<1nf3$ for electronic transitions between adjacent orbits.

Answer 15

(b)

Expert Solution

Answer to Problem 31P

It is showed that frequency of the photon emitted when an electron jumps from an outer to inner orbit can be written as $fphoton=mek2Z2e42πℏ3ni+nf2ni2nf2(ni−nf)$ and $1ni3<ni+nf2ni2nf2<1nf3$ for electronic transitions between adjacent orbits.

Answer 16

(b)

Expert Solution

Answer 17

(b)

Expert Solution

Answer 18

Expert Solution

Answer 19

Explanation of Solution

Write the equation for the energy of the photon emitted.

$ΔE=hfphoton$

Here, $ΔE$ is the energy of the photon emitted, $h$ is the Planck’s constant and $fphoton$ is the frequency of the photon.

Write the equation for the energy of the photon emitted when an electron jumps from an outer to inner orbit.

$ΔE=h(kZ2e22a0h)(1nf2−1ni2)$

Here, $ni$ is the quantum number of the initial level and $nf$ is the quantum number of the final level.

Compare the above two equations to write the expression for $fphoton$ .

$fphoton=(kZ2e22a0h)(1nf2−1ni2)=(kZ2e22a0h)(ni2−nf2ni2nf2)=(kZ2e22a0h)(ni+nf)(ni−nf)ni2nf2$

Put the equation for Bohr radius in the above equation.

$fphoton=(kZ2e22(ℏ2meke2)h)(ni+nf)(ni−nf)ni2nf2=mek2Z2e42ℏ2h(ni+nf)(ni−nf)ni2nf2=mek2Z2e42ℏ2(h2π)(2π)(ni+nf)(ni−nf)ni2nf2=mek2Z2e42πℏ3(ni+nf)(ni−nf)2ni2nf2$ (V)

Write the relationship between $ni$ and $nf$ for transitions between adjacent orbits.

$ni−nf=1$

Put the above equation in equation (V).

$fphoton=mek2Z2e42πℏ3(ni+nf)(1)2ni2nf2=mek2Z2e42πℏ3(ni+nf)2ni2nf2$ (VI)

Conclusion:

For $ni=2,nf=1,ni=3,nf=2, etc.,$ using equation (VI) gives that $1ni3<ni+nf2ni2nf2<1nf3$ .

Therefore, it is showed that frequency of the photon emitted when an electron jumps from an outer to inner orbit can be written as $fphoton=mek2Z2e42πℏ3ni+nf2ni2nf2(ni−nf)$ and $1ni3<ni+nf2ni2nf2<1nf3$ for electronic transitions between adjacent orbits.

Answer 20

(c)

To determine

The conclusion regarding the frequency of the emitted radiation compared with the frequencies of orbital revolution in the initial and final states and the result as $ni→∞$ .

(c)

Expert Solution

Answer to Problem 31P

Frequency of the emitted radiation is in between the initial and the final orbital frequency and the frequency of the emitted radiation becomes the orbital frequency as $ni→∞$ .

Explanation of Solution

The expression for the orbital frequency of the electron is found in part (a). The equation for the frequency of the emitted photon is found in part (b). Comparison of the two equations shows that the frequency of the emitted radiation is in between the initial orbital frequency and the final orbital frequency.

As $ni→∞$ , the initial orbital frequency and the final orbital frequency squeeze closer together. This will make the frequency of the emitted radiation equal to the orbital frequency. This result is in agreement with Bohr’s correspondence principle.

Conclusion:

Thus, the frequency of the emitted radiation is in between the initial and the final orbital frequency and the frequency of the emitted radiation becomes the orbital frequency as $ni→∞$ .

Answer 21

(c)

To determine

The conclusion regarding the frequency of the emitted radiation compared with the frequencies of orbital revolution in the initial and final states and the result as $ni→∞$ .

Answer 22

(c)

Expert Solution

Answer to Problem 31P

Frequency of the emitted radiation is in between the initial and the final orbital frequency and the frequency of the emitted radiation becomes the orbital frequency as $ni→∞$ .

Answer 23

(c)

Expert Solution

Answer 24

(c)

Expert Solution

Answer 25

Expert Solution

Answer 26

Explanation of Solution

The expression for the orbital frequency of the electron is found in part (a). The equation for the frequency of the emitted photon is found in part (b). Comparison of the two equations shows that the frequency of the emitted radiation is in between the initial orbital frequency and the final orbital frequency.

As $ni→∞$ , the initial orbital frequency and the final orbital frequency squeeze closer together. This will make the frequency of the emitted radiation equal to the orbital frequency. This result is in agreement with Bohr’s correspondence principle.

Conclusion:

Thus, the frequency of the emitted radiation is in between the initial and the final orbital frequency and the frequency of the emitted radiation becomes the orbital frequency as $ni→∞$ .

Answer 27

Ch. 4.2 - Exercise 1 Find the horizontal speed vx for this...Ch. 4.2 - Prob. 2E Ch. 4.3 - Prob. 3E Ch. 4.3 - Prob. 4E Ch. 4.3 - Prob. 5E Ch. 4 - Prob. 1Q Ch. 4 - Prob. 2Q Ch. 4 - Prob. 3Q Ch. 4 - Prob. 4Q Ch. 4 - Prob. 5Q

To show that the frequency of the electron’s orbital motion is f e = m e k 2 Z 2 e 4 2 π ℏ 3 ( 1 n 3 ) .

Concept explainers

To show that the frequency of the electron’s orbital motion is $fe=mek2Z2e42πℏ3(1n3)$ .

Answer to Problem 31P

Explanation of Solution

To show that frequency of the photon emitted when an electron jumps from an outer to inner orbit can be written as $fe=mek2Z2e42πℏ3ni+nf2ni2nf2(ni−nf)$ and $1ni3<ni+nf2ni2nf2<1nf3$ for electronic transitions between adjacent orbits.

Answer to Problem 31P

Explanation of Solution

The conclusion regarding the frequency of the emitted radiation compared with the frequencies of orbital revolution in the initial and final states and the result as $ni→∞$ .

Answer to Problem 31P

Explanation of Solution

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Chapter 4 Solutions

To show that the frequency of the electron’s orbital motion is f e = m e k 2 Z 2 e 4 2 π ℏ 3 ( 1 n 3 ) .

Concept explainers

Answer to Problem 31P

Explanation of Solution

Answer to Problem 31P

Explanation of Solution

The conclusion regarding the frequency of the emitted radiation compared with the frequencies of orbital revolution in the initial and final states and the result as ni→∞<m:math><m:mrow><m:msub><m:mi>n</m:mi><m:mi>i</m:mi></m:msub><m:mo>→</m:mo><m:mi>∞</m:mi></m:mrow></m:math> .

Answer to Problem 31P

Explanation of Solution

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Chapter 4 Solutions

The conclusion regarding the frequency of the emitted radiation compared with the frequencies of orbital revolution in the initial and final states and the result as $ni→∞$ .