Chapter 28, Problem 28P

(a)

To determine

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

Answer to Problem 28P

The expression for the energy level of the sole remaining electron is $E_n=-\frac{\left(122\text{eV}\right)}{n^2}$.

Explanation of Solution

Formula to calculate the energy level is,

  $E_n=-\frac{Z^2\left(13.6\text{eV}\right)}{n^2}$

• $E_n$ is the nth energy level,
• $n$ is nth level
• $Z$ is the atomic number

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

  $\begin{array}{c}{E}_n=-\frac{{\left(3\right)}^2\left(13.6\text{eV}\right)}{n^2}\\ =-\frac{\left(122\text{eV}\right)}{n^2}\end{array}$

Thus, expression for the energy level is $-\frac{\left(122\text{eV}\right)}{n^2}$.

Conclusion:

Therefore, the expression for the energy level is $-\frac{\left(122\text{eV}\right)}{n^2}$.

(b)

To determine

The energy for the level $n=4$.

Answer to Problem 28P

The energy for the level $n=4$ is $-7.63\text{eV}$.

Explanation of Solution

Formula to calculate the energy level is,

  $E_n=-\frac{\left(122\text{eV}\right)}{n^2}$

• $E_n$ is the nth energy level,
• $n$ is nth level

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

  $\begin{array}{c}{E}_4=-\frac{\left(122\text{eV}\right)}{{\left(4\right)}^2}\\ =-7.63\text{eV}\end{array}$

Thus, the energy for the level $n=4$ is $-7.63\text{eV}$.

Conclusion:

Therefore, the energy for the level $n=4$ is $-7.63\text{eV}$.

(c)

To determine

The energy for the level $n=2$.

Answer to Problem 28P

The energy for the level $n=2$ is $-30.5\text{eV}$.

Explanation of Solution

Formula to calculate the energy level is,

  $E_n=-\frac{\left(122\text{eV}\right)}{n^2}$

• $E_n$ is the nth energy level,
• $n$ is nth level

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

  $\begin{array}{c}{E}_2=-\frac{\left(122\text{eV}\right)}{{\left(2\right)}^2}\\ =-30.5\text{eV}\end{array}$

Thus, the energy for the level $n=2$ is $-30.5\text{eV}$.

Conclusion:

Therefore, the energy for the level $n=2$ is $-30.5\text{eV}$.

(d)

To determine

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

Answer to Problem 28P

the energy of the photon for the transition from fourth level to second level is $22.9\text{eV}$ or $3.66\times {10}^{-18}\text{J}$.

Explanation of Solution

Formula to calculate the energy difference is,

  $E_{photon}=\left(E_4-E_2\right)$

• $E_2\text{and}{E}_4$ are the second and fourth level energy,
• $E_n$ is nth level energy

From unit conversion,

    $1\text{eV}=\text{1}\text{.6}\times {\text{10}}^{-19}\text{J}$

Substitute $\left(-7.63\text{eV}\right)$ for $E_4$, $\left(-30.5\text{eV}\right)$ for $E_2$ to find $E_{photon}$.

  $\begin{array}{c}{E}_{photon}=\left[\left(-7.63\text{eV}\right)-\left(-30.5\text{eV}\right)\right]\\ =22.9\text{eV}\\ =22.9\text{eV}\times \frac{1.6\times {10}^{-19}\text{J}}{1\text{eV}}\\ =3.66\times {10}^{-18}\text{J}\end{array}$

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

$3.66\times {10}^{-18}\text{J}$.

Conclusion:

Therefore, the energy of the photon for the transition from fourth level to second level is $22.9\text{eV}$ or $3.66\times {10}^{-18}\text{J}$.

(e)

To determine

The frequency and wavelength of the emitted photon.

Answer to Problem 28P

The frequency and wavelength of the emitted photon is $5.52\times {10}^{15}\text{Hz}$ and $5.43\times {10}^{-8}\text{m}$ respectively.

Explanation of Solution

Formula to calculate the frequency of the photon is,

  $f=\frac{E_{photon}}{h}$

• $E_{photon}$ is the photon energy
• $h$ is Planck’s constant

Substitute $3.66\times {10}^{-18}\text{J}$ for $E_{photon}$, $6.63\times {10}^{-34}\text{J-s}$ for $h$ to find $f$.

  $\begin{array}{c}f=\frac{\left(3.66\times {10}^{-18}\text{J}\right)}{\left(6.63\times {10}^{-34}\text{J-s}\right)}\\ =5.52\times {10}^{15}\text{Hz}\end{array}$

Formula to calculate the wavelength of the photon is,

  $\lambda =\frac{c}{f}$

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

Substitute $3\times {10}^8\text{m/s}$ for $c$, $5.52\times {10}^{15}\text{Hz}$ for $f$ to find $\lambda$.

  $\begin{array}{c}\lambda =\frac{3\times {10}^8\text{m/s}}{5.52\times {10}^{15}\text{Hz}}\\ =5.43\times {10}^{-8}\text{m}\end{array}$

Thus, the frequency and wavelength of the emitted photon is $5.52\times {10}^{15}\text{Hz}$ and $5.43\times {10}^{-8}\text{m}$ respectively.

Conclusion:

Therefore, the frequency and wavelength of the emitted photon is $5.52\times {10}^{15}\text{Hz}$ and $5.43\times {10}^{-8}\text{m}$ respectively.

(f)

To determine

The wavelength belongs to in which spectrum.

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 $5.43\times {10}^{-8}\text{m}$. 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