The expression for the energy level of the sole remaining electron.

Question

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Answer 1

Question

Chapter 28, Problem 28P

(a)

To determine

The expression for the energy level of the sole remaining electron.

(a)

Expert Solution

Answer to Problem 28P

The expression for the energy level of the sole remaining electron is $En=−(122 eV)n2$ .

Explanation of Solution

Formula to calculate the energy level is,

$En=−Z2(13.6 eV)n2$

$En$ is the nth energy level,
$n$ is nth level
$Z$ is the atomic number

Substitute $3$ for $Z$ to find $En$ .

$En=−(3)2(13.6 eV)n2=−(122 eV)n2$

Thus, expression for the energy level is $−(122 eV)n2$ .

Conclusion:

Therefore, the expression for the energy level is $−(122 eV)n2$ .

(b)

To determine

The energy for the level $n=4$ .

(b)

Expert Solution

Answer to Problem 28P

The energy for the level $n=4$ is $−7.63 eV$ .

Explanation of Solution

Formula to calculate the energy level is,

$En=−(122 eV)n2$

$En$ is the nth energy level,
$n$ is nth level

Substitute $4$ for $n$ to find $En$ .

$E4=−(122 eV)(4)2=−7.63 eV$

Thus, the energy for the level $n=4$ is $−7.63 eV$ .

Conclusion:

Therefore, the energy for the level $n=4$ is $−7.63 eV$ .

(c)

To determine

The energy for the level $n=2$ .

(c)

Expert Solution

Answer to Problem 28P

The energy for the level $n=2$ is $−30.5 eV$ .

Explanation of Solution

Formula to calculate the energy level is,

$En=−(122 eV)n2$

$En$ is the nth energy level,
$n$ is nth level

Substitute $2$ for $n$ to find $En$ .

$E2=−(122 eV)(2)2=−30.5 eV$

Thus, the energy for the level $n=2$ is $−30.5 eV$ .

Conclusion:

Therefore, the energy for the level $n=2$ is $−30.5 eV$ .

(d)

To determine

The energy of the photon for the transition from fourth level to second level.

(d)

Expert Solution

Answer to Problem 28P

the energy of the photon for the transition from fourth level to second level is $22.9 eV$ or $3.66×10−18 J$ .

Explanation of Solution

Formula to calculate the energy difference is,

$Ephoton=(E4−E2)$

$E2 and E4$ are the second and fourth level energy,
$En$ is nth level energy

From unit conversion,

$1 eV=1.6×10−19J$

Substitute $(−7.63 eV)$ for $E4$ , $(−30.5 eV)$ for $E2$ to find $Ephoton$ .

$Ephoton=[(−7.63 eV)−(−30.5 eV)]=22.9 eV=22.9 eV×1.6×10−19J1 eV=3.66×10−18 J$

Thus, the energy of the photon for the transition from fourth level to second level is $22.9 eV$ or

$3.66×10−18 J$ .

Conclusion:

Therefore, the energy of the photon for the transition from fourth level to second level is $22.9 eV$ or $3.66×10−18 J$ .

(e)

To determine

The frequency and wavelength of the emitted photon.

(e)

Expert Solution

Answer to Problem 28P

The frequency and wavelength of the emitted photon is $5.52×1015 Hz$ and $5.43×10−8m$ respectively.

Explanation of Solution

Formula to calculate the frequency of the photon is,

$f=Ephotonh$

$Ephoton$ is the photon energy
$h$ is Planck’s constant

Substitute $3.66×10−18J$ for $Ephoton$ , $6.63×10−34 J-s$ for $h$ to find $f$ .

$f=(3.66×10−18 J)(6.63×10−34 J-s)=5.52×1015 Hz$

Formula to calculate the wavelength of the photon is,

$λ=cf$

$c$ is the speed of light
$f$ is the frequency

Substitute $3×108m/s$ for $c$ , $5.52×1015 Hz$ for $f$ to find $λ$ .

$λ=3×108 m/s5.52×1015 Hz=5.43×10−8m$

Thus, the frequency and wavelength of the emitted photon is $5.52×1015 Hz$ and $5.43×10−8m$ respectively.

Conclusion:

Therefore, the frequency and wavelength of the emitted photon is $5.52×1015 Hz$ and $5.43×10−8m$ respectively.

(f)

To determine

The wavelength belongs to in which spectrum.

(f)

Expert Solution

Answer to Problem 28P

The wavelength belongs to in which spectrum is deep ultraviolet region.

Explanation of Solution

The wavelength of the photon for the transition is $5.43×10−8m$ . So, this wavelength belongs to deep ultraviolet region.

Thus, this wavelength belongs to deep ultraviolet region.

Conclusion:

Therefore, the wavelength belongs to deep ultraviolet region

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Answer 2

Question

Chapter 28, Problem 28P

(a)

To determine

The expression for the energy level of the sole remaining electron.

(a)

Expert Solution

Answer to Problem 28P

The expression for the energy level of the sole remaining electron is $En=−(122 eV)n2$ .

Explanation of Solution

Formula to calculate the energy level is,

$En=−Z2(13.6 eV)n2$

$En$ is the nth energy level,
$n$ is nth level
$Z$ is the atomic number

Substitute $3$ for $Z$ to find $En$ .

$En=−(3)2(13.6 eV)n2=−(122 eV)n2$

Thus, expression for the energy level is $−(122 eV)n2$ .

Conclusion:

Therefore, the expression for the energy level is $−(122 eV)n2$ .

(b)

To determine

The energy for the level $n=4$ .

(b)

Expert Solution

Answer to Problem 28P

The energy for the level $n=4$ is $−7.63 eV$ .

Explanation of Solution

Formula to calculate the energy level is,

$En=−(122 eV)n2$

$En$ is the nth energy level,
$n$ is nth level

Substitute $4$ for $n$ to find $En$ .

$E4=−(122 eV)(4)2=−7.63 eV$

Thus, the energy for the level $n=4$ is $−7.63 eV$ .

Conclusion:

Therefore, the energy for the level $n=4$ is $−7.63 eV$ .

(c)

To determine

The energy for the level $n=2$ .

(c)

Expert Solution

Answer to Problem 28P

The energy for the level $n=2$ is $−30.5 eV$ .

Explanation of Solution

Formula to calculate the energy level is,

$En=−(122 eV)n2$

$En$ is the nth energy level,
$n$ is nth level

Substitute $2$ for $n$ to find $En$ .

$E2=−(122 eV)(2)2=−30.5 eV$

Thus, the energy for the level $n=2$ is $−30.5 eV$ .

Conclusion:

Therefore, the energy for the level $n=2$ is $−30.5 eV$ .

(d)

To determine

The energy of the photon for the transition from fourth level to second level.

(d)

Expert Solution

Answer to Problem 28P

the energy of the photon for the transition from fourth level to second level is $22.9 eV$ or $3.66×10−18 J$ .

Explanation of Solution

Formula to calculate the energy difference is,

$Ephoton=(E4−E2)$

$E2 and E4$ are the second and fourth level energy,
$En$ is nth level energy

From unit conversion,

$1 eV=1.6×10−19J$

Substitute $(−7.63 eV)$ for $E4$ , $(−30.5 eV)$ for $E2$ to find $Ephoton$ .

$Ephoton=[(−7.63 eV)−(−30.5 eV)]=22.9 eV=22.9 eV×1.6×10−19J1 eV=3.66×10−18 J$

Thus, the energy of the photon for the transition from fourth level to second level is $22.9 eV$ or

$3.66×10−18 J$ .

Conclusion:

Therefore, the energy of the photon for the transition from fourth level to second level is $22.9 eV$ or $3.66×10−18 J$ .

(e)

To determine

The frequency and wavelength of the emitted photon.

(e)

Expert Solution

Answer to Problem 28P

The frequency and wavelength of the emitted photon is $5.52×1015 Hz$ and $5.43×10−8m$ respectively.

Explanation of Solution

Formula to calculate the frequency of the photon is,

$f=Ephotonh$

$Ephoton$ is the photon energy
$h$ is Planck’s constant

Substitute $3.66×10−18J$ for $Ephoton$ , $6.63×10−34 J-s$ for $h$ to find $f$ .

$f=(3.66×10−18 J)(6.63×10−34 J-s)=5.52×1015 Hz$

Formula to calculate the wavelength of the photon is,

$λ=cf$

$c$ is the speed of light
$f$ is the frequency

Substitute $3×108m/s$ for $c$ , $5.52×1015 Hz$ for $f$ to find $λ$ .

$λ=3×108 m/s5.52×1015 Hz=5.43×10−8m$

Thus, the frequency and wavelength of the emitted photon is $5.52×1015 Hz$ and $5.43×10−8m$ respectively.

Conclusion:

Therefore, the frequency and wavelength of the emitted photon is $5.52×1015 Hz$ and $5.43×10−8m$ respectively.

(f)

To determine

The wavelength belongs to in which spectrum.

(f)

Expert Solution

Answer to Problem 28P

The wavelength belongs to in which spectrum is deep ultraviolet region.

Explanation of Solution

The wavelength of the photon for the transition is $5.43×10−8m$ . So, this wavelength belongs to deep ultraviolet region.

Thus, this wavelength belongs to deep ultraviolet region.

Conclusion:

Therefore, the wavelength belongs to deep ultraviolet region

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Answer 3

Question

Chapter 28, Problem 28P

(a)

To determine

The expression for the energy level of the sole remaining electron.

(a)

Expert Solution

Answer to Problem 28P

The expression for the energy level of the sole remaining electron is $En=−(122 eV)n2$ .

Explanation of Solution

Formula to calculate the energy level is,

$En=−Z2(13.6 eV)n2$

$En$ is the nth energy level,
$n$ is nth level
$Z$ is the atomic number

Substitute $3$ for $Z$ to find $En$ .

$En=−(3)2(13.6 eV)n2=−(122 eV)n2$

Thus, expression for the energy level is $−(122 eV)n2$ .

Conclusion:

Therefore, the expression for the energy level is $−(122 eV)n2$ .

(b)

To determine

The energy for the level $n=4$ .

(b)

Expert Solution

Answer to Problem 28P

The energy for the level $n=4$ is $−7.63 eV$ .

Explanation of Solution

Formula to calculate the energy level is,

$En=−(122 eV)n2$

$En$ is the nth energy level,
$n$ is nth level

Substitute $4$ for $n$ to find $En$ .

$E4=−(122 eV)(4)2=−7.63 eV$

Thus, the energy for the level $n=4$ is $−7.63 eV$ .

Conclusion:

Therefore, the energy for the level $n=4$ is $−7.63 eV$ .

(c)

To determine

The energy for the level $n=2$ .

(c)

Expert Solution

Answer to Problem 28P

The energy for the level $n=2$ is $−30.5 eV$ .

Explanation of Solution

Formula to calculate the energy level is,

$En=−(122 eV)n2$

$En$ is the nth energy level,
$n$ is nth level

Substitute $2$ for $n$ to find $En$ .

$E2=−(122 eV)(2)2=−30.5 eV$

Thus, the energy for the level $n=2$ is $−30.5 eV$ .

Conclusion:

Therefore, the energy for the level $n=2$ is $−30.5 eV$ .

(d)

To determine

The energy of the photon for the transition from fourth level to second level.

(d)

Expert Solution

Answer to Problem 28P

the energy of the photon for the transition from fourth level to second level is $22.9 eV$ or $3.66×10−18 J$ .

Explanation of Solution

Formula to calculate the energy difference is,

$Ephoton=(E4−E2)$

$E2 and E4$ are the second and fourth level energy,
$En$ is nth level energy

From unit conversion,

$1 eV=1.6×10−19J$

Substitute $(−7.63 eV)$ for $E4$ , $(−30.5 eV)$ for $E2$ to find $Ephoton$ .

$Ephoton=[(−7.63 eV)−(−30.5 eV)]=22.9 eV=22.9 eV×1.6×10−19J1 eV=3.66×10−18 J$

Thus, the energy of the photon for the transition from fourth level to second level is $22.9 eV$ or

$3.66×10−18 J$ .

Conclusion:

Therefore, the energy of the photon for the transition from fourth level to second level is $22.9 eV$ or $3.66×10−18 J$ .

(e)

To determine

The frequency and wavelength of the emitted photon.

(e)

Expert Solution

Answer to Problem 28P

The frequency and wavelength of the emitted photon is $5.52×1015 Hz$ and $5.43×10−8m$ respectively.

Explanation of Solution

Formula to calculate the frequency of the photon is,

$f=Ephotonh$

$Ephoton$ is the photon energy
$h$ is Planck’s constant

Substitute $3.66×10−18J$ for $Ephoton$ , $6.63×10−34 J-s$ for $h$ to find $f$ .

$f=(3.66×10−18 J)(6.63×10−34 J-s)=5.52×1015 Hz$

Formula to calculate the wavelength of the photon is,

$λ=cf$

$c$ is the speed of light
$f$ is the frequency

Substitute $3×108m/s$ for $c$ , $5.52×1015 Hz$ for $f$ to find $λ$ .

$λ=3×108 m/s5.52×1015 Hz=5.43×10−8m$

Thus, the frequency and wavelength of the emitted photon is $5.52×1015 Hz$ and $5.43×10−8m$ respectively.

Conclusion:

Therefore, the frequency and wavelength of the emitted photon is $5.52×1015 Hz$ and $5.43×10−8m$ respectively.

(f)

To determine

The wavelength belongs to in which spectrum.

(f)

Expert Solution

Answer to Problem 28P

The wavelength belongs to in which spectrum is deep ultraviolet region.

Explanation of Solution

The wavelength of the photon for the transition is $5.43×10−8m$ . So, this wavelength belongs to deep ultraviolet region.

Thus, this wavelength belongs to deep ultraviolet region.

Conclusion:

Therefore, the wavelength belongs to deep ultraviolet region

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Answer 4

Question

Chapter 28, Problem 28P

(a)

To determine

The expression for the energy level of the sole remaining electron.

(a)

Expert Solution

Answer to Problem 28P

The expression for the energy level of the sole remaining electron is $En=−(122 eV)n2$ .

Explanation of Solution

Formula to calculate the energy level is,

$En=−Z2(13.6 eV)n2$

$En$ is the nth energy level,
$n$ is nth level
$Z$ is the atomic number

Substitute $3$ for $Z$ to find $En$ .

$En=−(3)2(13.6 eV)n2=−(122 eV)n2$

Thus, expression for the energy level is $−(122 eV)n2$ .

Conclusion:

Therefore, the expression for the energy level is $−(122 eV)n2$ .

(b)

To determine

The energy for the level $n=4$ .

(b)

Expert Solution

Answer to Problem 28P

The energy for the level $n=4$ is $−7.63 eV$ .

Explanation of Solution

Formula to calculate the energy level is,

$En=−(122 eV)n2$

$En$ is the nth energy level,
$n$ is nth level

Substitute $4$ for $n$ to find $En$ .

$E4=−(122 eV)(4)2=−7.63 eV$

Thus, the energy for the level $n=4$ is $−7.63 eV$ .

Conclusion:

Therefore, the energy for the level $n=4$ is $−7.63 eV$ .

(c)

To determine

The energy for the level $n=2$ .

(c)

Expert Solution

Answer to Problem 28P

The energy for the level $n=2$ is $−30.5 eV$ .

Explanation of Solution

Formula to calculate the energy level is,

$En=−(122 eV)n2$

$En$ is the nth energy level,
$n$ is nth level

Substitute $2$ for $n$ to find $En$ .

$E2=−(122 eV)(2)2=−30.5 eV$

Thus, the energy for the level $n=2$ is $−30.5 eV$ .

Conclusion:

Therefore, the energy for the level $n=2$ is $−30.5 eV$ .

(d)

To determine

The energy of the photon for the transition from fourth level to second level.

(d)

Expert Solution

Answer to Problem 28P

the energy of the photon for the transition from fourth level to second level is $22.9 eV$ or $3.66×10−18 J$ .

Explanation of Solution

Formula to calculate the energy difference is,

$Ephoton=(E4−E2)$

$E2 and E4$ are the second and fourth level energy,
$En$ is nth level energy

From unit conversion,

$1 eV=1.6×10−19J$

Substitute $(−7.63 eV)$ for $E4$ , $(−30.5 eV)$ for $E2$ to find $Ephoton$ .

$Ephoton=[(−7.63 eV)−(−30.5 eV)]=22.9 eV=22.9 eV×1.6×10−19J1 eV=3.66×10−18 J$

Thus, the energy of the photon for the transition from fourth level to second level is $22.9 eV$ or

$3.66×10−18 J$ .

Conclusion:

Therefore, the energy of the photon for the transition from fourth level to second level is $22.9 eV$ or $3.66×10−18 J$ .

(e)

To determine

The frequency and wavelength of the emitted photon.

(e)

Expert Solution

Answer to Problem 28P

The frequency and wavelength of the emitted photon is $5.52×1015 Hz$ and $5.43×10−8m$ respectively.

Explanation of Solution

Formula to calculate the frequency of the photon is,

$f=Ephotonh$

$Ephoton$ is the photon energy
$h$ is Planck’s constant

Substitute $3.66×10−18J$ for $Ephoton$ , $6.63×10−34 J-s$ for $h$ to find $f$ .

$f=(3.66×10−18 J)(6.63×10−34 J-s)=5.52×1015 Hz$

Formula to calculate the wavelength of the photon is,

$λ=cf$

$c$ is the speed of light
$f$ is the frequency

Substitute $3×108m/s$ for $c$ , $5.52×1015 Hz$ for $f$ to find $λ$ .

$λ=3×108 m/s5.52×1015 Hz=5.43×10−8m$

Thus, the frequency and wavelength of the emitted photon is $5.52×1015 Hz$ and $5.43×10−8m$ respectively.

Conclusion:

Therefore, the frequency and wavelength of the emitted photon is $5.52×1015 Hz$ and $5.43×10−8m$ respectively.

(f)

To determine

The wavelength belongs to in which spectrum.

(f)

Expert Solution

Answer to Problem 28P

The wavelength belongs to in which spectrum is deep ultraviolet region.

Explanation of Solution

The wavelength of the photon for the transition is $5.43×10−8m$ . So, this wavelength belongs to deep ultraviolet region.

Thus, this wavelength belongs to deep ultraviolet region.

Conclusion:

Therefore, the wavelength belongs to deep ultraviolet region

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Answer 5

Chapter 28, Problem 28P

(a)

To determine

The expression for the energy level of the sole remaining electron.

(a)

Expert Solution

Answer to Problem 28P

The expression for the energy level of the sole remaining electron is $En=−(122 eV)n2$ .

Explanation of Solution

Formula to calculate the energy level is,

$En=−Z2(13.6 eV)n2$

$En$ is the nth energy level,
$n$ is nth level
$Z$ is the atomic number

Substitute $3$ for $Z$ to find $En$ .

$En=−(3)2(13.6 eV)n2=−(122 eV)n2$

Thus, expression for the energy level is $−(122 eV)n2$ .

Conclusion:

Therefore, the expression for the energy level is $−(122 eV)n2$ .

(b)

To determine

The energy for the level $n=4$ .

(b)

Expert Solution

Answer to Problem 28P

The energy for the level $n=4$ is $−7.63 eV$ .

Explanation of Solution

Formula to calculate the energy level is,

$En=−(122 eV)n2$

$En$ is the nth energy level,
$n$ is nth level

Substitute $4$ for $n$ to find $En$ .

$E4=−(122 eV)(4)2=−7.63 eV$

Thus, the energy for the level $n=4$ is $−7.63 eV$ .

Conclusion:

Therefore, the energy for the level $n=4$ is $−7.63 eV$ .

(c)

To determine

The energy for the level $n=2$ .

(c)

Expert Solution

Answer to Problem 28P

The energy for the level $n=2$ is $−30.5 eV$ .

Explanation of Solution

Formula to calculate the energy level is,

$En=−(122 eV)n2$

$En$ is the nth energy level,
$n$ is nth level

Substitute $2$ for $n$ to find $En$ .

$E2=−(122 eV)(2)2=−30.5 eV$

Thus, the energy for the level $n=2$ is $−30.5 eV$ .

Conclusion:

Therefore, the energy for the level $n=2$ is $−30.5 eV$ .

(d)

To determine

The energy of the photon for the transition from fourth level to second level.

(d)

Expert Solution

Answer to Problem 28P

the energy of the photon for the transition from fourth level to second level is $22.9 eV$ or $3.66×10−18 J$ .

Explanation of Solution

Formula to calculate the energy difference is,

$Ephoton=(E4−E2)$

$E2 and E4$ are the second and fourth level energy,
$En$ is nth level energy

From unit conversion,

$1 eV=1.6×10−19J$

Substitute $(−7.63 eV)$ for $E4$ , $(−30.5 eV)$ for $E2$ to find $Ephoton$ .

$Ephoton=[(−7.63 eV)−(−30.5 eV)]=22.9 eV=22.9 eV×1.6×10−19J1 eV=3.66×10−18 J$

Thus, the energy of the photon for the transition from fourth level to second level is $22.9 eV$ or

$3.66×10−18 J$ .

Conclusion:

Therefore, the energy of the photon for the transition from fourth level to second level is $22.9 eV$ or $3.66×10−18 J$ .

(e)

To determine

The frequency and wavelength of the emitted photon.

(e)

Expert Solution

Answer to Problem 28P

The frequency and wavelength of the emitted photon is $5.52×1015 Hz$ and $5.43×10−8m$ respectively.

Explanation of Solution

Formula to calculate the frequency of the photon is,

$f=Ephotonh$

$Ephoton$ is the photon energy
$h$ is Planck’s constant

Substitute $3.66×10−18J$ for $Ephoton$ , $6.63×10−34 J-s$ for $h$ to find $f$ .

$f=(3.66×10−18 J)(6.63×10−34 J-s)=5.52×1015 Hz$

Formula to calculate the wavelength of the photon is,

$λ=cf$

$c$ is the speed of light
$f$ is the frequency

Substitute $3×108m/s$ for $c$ , $5.52×1015 Hz$ for $f$ to find $λ$ .

$λ=3×108 m/s5.52×1015 Hz=5.43×10−8m$

Thus, the frequency and wavelength of the emitted photon is $5.52×1015 Hz$ and $5.43×10−8m$ respectively.

Conclusion:

Therefore, the frequency and wavelength of the emitted photon is $5.52×1015 Hz$ and $5.43×10−8m$ respectively.

(f)

To determine

The wavelength belongs to in which spectrum.

(f)

Expert Solution

Answer to Problem 28P

The wavelength belongs to in which spectrum is deep ultraviolet region.

Explanation of Solution

The wavelength of the photon for the transition is $5.43×10−8m$ . So, this wavelength belongs to deep ultraviolet region.

Thus, this wavelength belongs to deep ultraviolet region.

Conclusion:

Therefore, the wavelength belongs to deep ultraviolet region

Answer 6

Chapter 28, Problem 28P

(a)

To determine

The expression for the energy level of the sole remaining electron.

(a)

Expert Solution

Answer to Problem 28P

The expression for the energy level of the sole remaining electron is $En=−(122 eV)n2$ .

Explanation of Solution

Formula to calculate the energy level is,

$En=−Z2(13.6 eV)n2$

$En$ is the nth energy level,
$n$ is nth level
$Z$ is the atomic number

Substitute $3$ for $Z$ to find $En$ .

$En=−(3)2(13.6 eV)n2=−(122 eV)n2$

Thus, expression for the energy level is $−(122 eV)n2$ .

Conclusion:

Therefore, the expression for the energy level is $−(122 eV)n2$ .

Answer 7

Chapter 28, Problem 28P

(a)

To determine

The expression for the energy level of the sole remaining electron.

Answer 8

(a)

Expert Solution

Answer to Problem 28P

The expression for the energy level of the sole remaining electron is $En=−(122 eV)n2$ .

Answer 9

(a)

Expert Solution

Answer 10

(a)

Expert Solution

Answer 11

Expert Solution

Answer 12

Explanation of Solution

Formula to calculate the energy level is,

$En=−Z2(13.6 eV)n2$

$En$ is the nth energy level,
$n$ is nth level
$Z$ is the atomic number

Substitute $3$ for $Z$ to find $En$ .

$En=−(3)2(13.6 eV)n2=−(122 eV)n2$

Thus, expression for the energy level is $−(122 eV)n2$ .

Conclusion:

Therefore, the expression for the energy level is $−(122 eV)n2$ .

Answer 13

(b)

To determine

The energy for the level $n=4$ .

(b)

Expert Solution

Answer to Problem 28P

The energy for the level $n=4$ is $−7.63 eV$ .

Explanation of Solution

Formula to calculate the energy level is,

$En=−(122 eV)n2$

$En$ is the nth energy level,
$n$ is nth level

Substitute $4$ for $n$ to find $En$ .

$E4=−(122 eV)(4)2=−7.63 eV$

Thus, the energy for the level $n=4$ is $−7.63 eV$ .

Conclusion:

Therefore, the energy for the level $n=4$ is $−7.63 eV$ .

Answer 14

(b)

To determine

The energy for the level $n=4$ .

Answer 15

(b)

Expert Solution

Answer to Problem 28P

The energy for the level $n=4$ is $−7.63 eV$ .

Answer 16

(b)

Expert Solution

Answer 17

(b)

Expert Solution

Answer 18

Expert Solution

Answer 19

Explanation of Solution

Formula to calculate the energy level is,

$En=−(122 eV)n2$

$En$ is the nth energy level,
$n$ is nth level

Substitute $4$ for $n$ to find $En$ .

$E4=−(122 eV)(4)2=−7.63 eV$

Thus, the energy for the level $n=4$ is $−7.63 eV$ .

Conclusion:

Therefore, the energy for the level $n=4$ is $−7.63 eV$ .

Answer 20

(c)

To determine

The energy for the level $n=2$ .

(c)

Expert Solution

Answer to Problem 28P

The energy for the level $n=2$ is $−30.5 eV$ .

Explanation of Solution

Formula to calculate the energy level is,

$En=−(122 eV)n2$

$En$ is the nth energy level,
$n$ is nth level

Substitute $2$ for $n$ to find $En$ .

$E2=−(122 eV)(2)2=−30.5 eV$

Thus, the energy for the level $n=2$ is $−30.5 eV$ .

Conclusion:

Therefore, the energy for the level $n=2$ is $−30.5 eV$ .

Answer 21

(c)

To determine

The energy for the level $n=2$ .

Answer 22

(c)

Expert Solution

Answer to Problem 28P

The energy for the level $n=2$ is $−30.5 eV$ .

Answer 23

(c)

Expert Solution

Answer 24

(c)

Expert Solution

Answer 25

Expert Solution

Answer 26

Explanation of Solution

Formula to calculate the energy level is,

$En=−(122 eV)n2$

$En$ is the nth energy level,
$n$ is nth level

Substitute $2$ for $n$ to find $En$ .

$E2=−(122 eV)(2)2=−30.5 eV$

Thus, the energy for the level $n=2$ is $−30.5 eV$ .

Conclusion:

Therefore, the energy for the level $n=2$ is $−30.5 eV$ .

Answer 27

(d)

To determine

The energy of the photon for the transition from fourth level to second level.

(d)

Expert Solution

Answer to Problem 28P

the energy of the photon for the transition from fourth level to second level is $22.9 eV$ or $3.66×10−18 J$ .

Explanation of Solution

Formula to calculate the energy difference is,

$Ephoton=(E4−E2)$

$E2 and E4$ are the second and fourth level energy,
$En$ is nth level energy

From unit conversion,

$1 eV=1.6×10−19J$

Substitute $(−7.63 eV)$ for $E4$ , $(−30.5 eV)$ for $E2$ to find $Ephoton$ .

$Ephoton=[(−7.63 eV)−(−30.5 eV)]=22.9 eV=22.9 eV×1.6×10−19J1 eV=3.66×10−18 J$

Thus, the energy of the photon for the transition from fourth level to second level is $22.9 eV$ or

$3.66×10−18 J$ .

Conclusion:

Therefore, the energy of the photon for the transition from fourth level to second level is $22.9 eV$ or $3.66×10−18 J$ .

Answer 28

(d)

To determine

The energy of the photon for the transition from fourth level to second level.

Answer 29

(d)

Expert Solution

Answer to Problem 28P

the energy of the photon for the transition from fourth level to second level is $22.9 eV$ or $3.66×10−18 J$ .

Answer 30

(d)

Expert Solution

Answer 31

(d)

Expert Solution

Answer 32

Expert Solution

Answer 33

Explanation of Solution

Formula to calculate the energy difference is,

$Ephoton=(E4−E2)$

$E2 and E4$ are the second and fourth level energy,
$En$ is nth level energy

From unit conversion,

$1 eV=1.6×10−19J$

Substitute $(−7.63 eV)$ for $E4$ , $(−30.5 eV)$ for $E2$ to find $Ephoton$ .

$Ephoton=[(−7.63 eV)−(−30.5 eV)]=22.9 eV=22.9 eV×1.6×10−19J1 eV=3.66×10−18 J$

Thus, the energy of the photon for the transition from fourth level to second level is $22.9 eV$ or

$3.66×10−18 J$ .

Conclusion:

Therefore, the energy of the photon for the transition from fourth level to second level is $22.9 eV$ or $3.66×10−18 J$ .

Answer 34

(e)

To determine

The frequency and wavelength of the emitted photon.

(e)

Expert Solution

Answer to Problem 28P

The frequency and wavelength of the emitted photon is $5.52×1015 Hz$ and $5.43×10−8m$ respectively.

Explanation of Solution

Formula to calculate the frequency of the photon is,

$f=Ephotonh$

$Ephoton$ is the photon energy
$h$ is Planck’s constant

Substitute $3.66×10−18J$ for $Ephoton$ , $6.63×10−34 J-s$ for $h$ to find $f$ .

$f=(3.66×10−18 J)(6.63×10−34 J-s)=5.52×1015 Hz$

Formula to calculate the wavelength of the photon is,

$λ=cf$

$c$ is the speed of light
$f$ is the frequency

Substitute $3×108m/s$ for $c$ , $5.52×1015 Hz$ for $f$ to find $λ$ .

$λ=3×108 m/s5.52×1015 Hz=5.43×10−8m$

Thus, the frequency and wavelength of the emitted photon is $5.52×1015 Hz$ and $5.43×10−8m$ respectively.

Conclusion:

Therefore, the frequency and wavelength of the emitted photon is $5.52×1015 Hz$ and $5.43×10−8m$ respectively.

Answer 35

(e)

To determine

The frequency and wavelength of the emitted photon.

Answer 36

(e)

Expert Solution

Answer to Problem 28P

The frequency and wavelength of the emitted photon is $5.52×1015 Hz$ and $5.43×10−8m$ respectively.

Answer 37

(e)

Expert Solution

Answer 38

(e)

Expert Solution

Answer 39

Expert Solution

Answer 40

Explanation of Solution

Formula to calculate the frequency of the photon is,

$f=Ephotonh$

$Ephoton$ is the photon energy
$h$ is Planck’s constant

Substitute $3.66×10−18J$ for $Ephoton$ , $6.63×10−34 J-s$ for $h$ to find $f$ .

$f=(3.66×10−18 J)(6.63×10−34 J-s)=5.52×1015 Hz$

Formula to calculate the wavelength of the photon is,

$λ=cf$

$c$ is the speed of light
$f$ is the frequency

Substitute $3×108m/s$ for $c$ , $5.52×1015 Hz$ for $f$ to find $λ$ .

$λ=3×108 m/s5.52×1015 Hz=5.43×10−8m$

Thus, the frequency and wavelength of the emitted photon is $5.52×1015 Hz$ and $5.43×10−8m$ respectively.

Conclusion:

Therefore, the frequency and wavelength of the emitted photon is $5.52×1015 Hz$ and $5.43×10−8m$ respectively.

Answer 41

(f)

To determine

The wavelength belongs to in which spectrum.

(f)

Expert Solution

Answer to Problem 28P

The wavelength belongs to in which spectrum is deep ultraviolet region.

Explanation of Solution

The wavelength of the photon for the transition is $5.43×10−8m$ . So, this wavelength belongs to deep ultraviolet region.

Thus, this wavelength belongs to deep ultraviolet region.

Conclusion:

Therefore, the wavelength belongs to deep ultraviolet region

Answer 42

(f)

To determine

The wavelength belongs to in which spectrum.

Answer 43

(f)

Expert Solution

Answer to Problem 28P

The wavelength belongs to in which spectrum is deep ultraviolet region.

Answer 44

(f)

Expert Solution

Answer 45

(f)

Expert Solution

Answer 46

Expert Solution

Answer 47

Explanation of Solution

The wavelength of the photon for the transition is $5.43×10−8m$ . So, this wavelength belongs to deep ultraviolet region.

Thus, this wavelength belongs to deep ultraviolet region.

Conclusion:

Therefore, the wavelength belongs to deep ultraviolet region

Answer 48

Ch. 28.3 - Prob. 28.1QQ Ch. 28.4 - Prob. 28.2QQ Ch. 28.5 - Prob. 28.3QQ Ch. 28 - Prob. 1CQ Ch. 28 - Prob. 2CQ Ch. 28 - Prob. 3CQ Ch. 28 - Prob. 4CQ Ch. 28 - Prob. 5CQ Ch. 28 - Prob. 6CQ Ch. 28 - Prob. 7CQ

The expression for the energy level of the sole remaining electron.

Concept explainers

The expression for the energy level of the sole remaining electron.

Answer to Problem 28P

Explanation of Solution

The energy for the level $n=4$ .

Answer to Problem 28P

Explanation of Solution

The energy for the level $n=2$ .

Answer to Problem 28P

Explanation of Solution

The energy of the photon for the transition from fourth level to second level.

Answer to Problem 28P

Explanation of Solution

The frequency and wavelength of the emitted photon.

Answer to Problem 28P

Explanation of Solution

The wavelength belongs to in which spectrum.

Answer to Problem 28P

Explanation of Solution

Want to see more full solutions like this?

Chapter 28 Solutions

The expression for the energy level of the sole remaining electron.

Concept explainers

The expression for the energy level of the sole remaining electron.

Answer to Problem 28P

Explanation of Solution

The energy for the level n=4<m:math><m:mrow><m:mi>n</m:mi><m:mo>=</m:mo><m:mn>4</m:mn></m:mrow></m:math>.

Answer to Problem 28P

Explanation of Solution

The energy for the level n=2<m:math><m:mrow><m:mi>n</m:mi><m:mo>=</m:mo><m:mn>2</m:mn></m:mrow></m:math>.

Answer to Problem 28P

Explanation of Solution

The energy of the photon for the transition from fourth level to second level.

Answer to Problem 28P

Explanation of Solution

The frequency and wavelength of the emitted photon.

Answer to Problem 28P

Explanation of Solution

The wavelength belongs to in which spectrum.

Answer to Problem 28P

Explanation of Solution

Want to see more full solutions like this?

Chapter 28 Solutions

The energy for the level $n=4$ .

The energy for the level $n=2$ .