Chapter 29, Problem 6P

(a)

To determine

The longest wavelength corresponding to a transition of photon’s energy.

Answer to Problem 6P

The longest wavelength corresponding to a transition of photon’s energy is $1.89\text{eV}$.

Explanation of Solution

The longest wavelength of the photon implies lowest frequency and smallest energy. The electron makes a transition from $n=3$ to $n=2$.

Write the expression for energy emitted by the electron in a transition from $n=3$ to $n=2$.

  $\Delta E=-\frac{13.6\text{eV}}{n^2}+\frac{13.6\text{eV}}{n^2}$

Here, $\Delta E$ is the energy and $n$ is the transition state of electron.

Conclusion:

The energy emitted by the electron in a transition from $n=3$ to $n=2$.

  $\begin{array}{c}\Delta E=-\frac{13.6\text{eV}}{3^2}+\frac{13.6\text{eV}}{2^2}\\ =1.89\text{eV}\end{array}$

Therefore, the longest wavelength corresponding to a transition of photon’s energy is $1.89\text{eV}$.

(b)

To determine

The longest wavelength of the corresponding transition.

Answer to Problem 6P

The longest wavelength of the corresponding transition is $656\text{nm}$.

Explanation of Solution

Write the expression from the relation between frequency and wavelength.

  $\lambda =\frac{c}{f}$        (I)

Here, $\lambda$ is the longest wavelength of the photon, $c$ is the velocity of light, and $f$ is the frequency.

Write the expression for photon’s energy.

  $\Delta E=hf$        (II)

Here, $\Delta E$ is the photon’s energy between the corresponding transitions and $h$ is the plank’s constant.

Rewrite the equation (II) for frequency.

  $f=\frac{\Delta E}{h}$        (III)

Conclusion:

Substitute equation (III) in equation (I).

  $\lambda =\frac{hc}{\Delta E}$

Substitute $6.626\times {10}^{-34}\text{J}\cdot \text{s}$ for $h$, $3\times {10}^8\text{m/s}$ for $c$, and $1.89\text{eV}$ for $\Delta E$ in the above equation to find $\lambda$.

  $\begin{array}{c}\lambda =\frac{\left(6.626\times {10}^{-34}\text{J}\cdot \text{s}\right)\left(3\times {10}^8\text{m/s}\right)}{\left(1.89\text{eV}\right)}\\ =\frac{19.878\times {10}^{-26}\text{J}\cdot \text{m}\left(\frac{6.242\times {10}^{18}\text{eV}}{1\text{J}}\right)}{\left(1.89\text{eV}\right)}\\ =\frac{1240\text{eV}\cdot \text{nm}}{1.89\text{eV}}\\ =656\text{nm}\end{array}$

This is the red Balmer alpha line, which gives its characteristic color to the chromospheres of the sun and to photograph of the Orion nebula.

Therefore, the longest wavelength of the corresponding transition is $656\text{nm}$.

(c)

To determine

The shortest wavelength corresponding to a transition of photon’s energy.

Answer to Problem 6P

The shortest wavelength corresponding to a transition of photon’s energy is $3.40\text{eV}$.

Explanation of Solution

The shortest wavelength of the photon implies highest frequency and greatest energy. The electron makes a transition from $n=\infty$ to $n=2$.

Write the expression for energy emitted by the electron in a transition from $n=\infty$ to $n=2$.

  $\Delta E=-\frac{13.6\text{eV}}{n^2}+\frac{13.6\text{eV}}{n^2}$

Here, $\Delta E$ is the energy and $n$ is the transition state of electron.

Conclusion:

The energy emitted by the electron in a transition from $n=\infty$ to $n=2$.

  $\begin{array}{c}\Delta E=-\frac{13.6\text{eV}}{\infty }+\frac{13.6\text{eV}}{2^2}\\ =3.40\text{eV}\end{array}$

Therefore, the shortest wavelength corresponding to a transition of photon’s energy is $3.40\text{eV}$.

(d)

To determine

The shortest wavelength of the corresponding transition.

Answer to Problem 6P

The shortest wavelength of the corresponding transition is $365\text{nm}$.

Explanation of Solution

Write the expression from the relation between frequency and wavelength.

  $\lambda =\frac{c}{f}$

Here, $\lambda$ is the shortest wavelength of the photon, $c$ is the velocity of light, and $f$ is the frequency.

Write the expression for photon’s energy.

  $\Delta E=hf$

Here, $\Delta E$ is the photon’s energy between the corresponding transitions and $h$ is the plank’s constant.

Rewrite the equation (II) for frequency.

  $f=\frac{\Delta E}{h}$

Conclusion:

Substitute equation (III) in equation (I).

  $\lambda =\frac{hc}{\Delta E}$

Substitute $6.626\times {10}^{-34}\text{J}\cdot \text{s}$ for $h$, $3\times {10}^8\text{m/s}$ for $c$, and $3.40\text{eV}$ for $\Delta E$ in the above equation to find $\lambda$.

  $\begin{array}{c}\lambda =\frac{\left(6.626\times {10}^{-34}\text{J}\cdot \text{s}\right)\left(3\times {10}^8\text{m/s}\right)}{\left(3.40\text{eV}\right)}\\ =\frac{19.878\times {10}^{-26}\text{J}\cdot \text{m}\left(\frac{6.242\times {10}^{18}\text{eV}}{1\text{J}}\right)}{\left(3.40\text{eV}\right)}\\ =\frac{1240\text{eV}\cdot \text{nm}}{3.40\text{eV}}\\ =365\text{nm}\end{array}$

Therefore, the smallest wavelength of the corresponding transition is $365\text{nm}$.

(e)

To determine

The shortest possible wavelength in the Balmer series.

Answer to Problem 6P

The shortest possible wavelength in the Balmer series is $365\text{nm}$.

Explanation of Solution

The transition limit in Balmer series is from $n=\infty$ to $n=2$, and their shortest possible wavelength is $365\text{nm}$, it is nearly equal to ultraviolet.

Conclusion:

Therefore, the shortest possible wavelength in the Balmer series is $365\text{nm}$.