Chapter 2.9, Problem 41AAP

(a)

To determine

The energy of the photon emitted.

Answer to Problem 41AAP

The energy of the photon emitted is $1.06\times {10}^{-19}\text{J}$.

Explanation of Solution

Write the expression to calculate energy of the photon emitted $\left(E\right)$.

 $\begin{array}{c}E=\Delta E\\ =\Delta \left(\frac{-13.6}{n^2}\right)\\ =\left(\frac{-13.6}{n_2^2}\right)-\left(\frac{-13.6}{n_1^2}\right)\end{array}$                                                                                        (I)

Here, change in energy of the photon between the states is $\Delta E$, principal quantum number is n and principal quantum numbers for state 3 and 4 are $n_1$ and $n_2$ respectively.

Conclusion:

Substitute $4$ for $n_2$ and $3$ for $n_1$ in Equation (I).

 $\begin{array}{c}E=\left(\frac{-13.6}{4^2}\right)-\left(\frac{-13.6}{3^2}\right)\\ =0.66\text{eV}\\ =0.66\text{eV}\times \frac{1.602\times {10}^{-19}\text{J}}{1\text{eV}}\\ =1.06\times {10}^{-19}\text{J}\end{array}$

Thus, the energy of the photon emitted is $1.06\times {10}^{-19}\text{J}$.

(b)

To determine

The frequency of the photon.

Answer to Problem 41AAP

The frequency of the photon is $1.6\times {10}^{14}\text{Hz}$.

Explanation of Solution

Write the expression to calculate frequency of the photon $\left(\upsilon \right)$.

 $\upsilon =\frac{E}{h}$                                                                                                                   (II)

Here, Planck's constant is $h$.

Conclusion:

The values of Planck constant is $6.63\times {10}^{-34}\text{J}\cdot \text{s}$.

Substitute $1.06\times {10}^{-19}\text{J}$ for $E$ and $6.63\times {10}^{-34}\text{J}\cdot \text{s}$ for $h$ in Equation (II).

 $\begin{array}{c}\upsilon =\frac{1.06\times {10}^{-19}\text{J}}{6.63\times {10}^{-34}\text{J}\cdot \text{s}}\\ =1.6\times {10}^{14}\text{Hz}\end{array}$

Thus, the frequency of the photon is $1.6\times {10}^{14}\text{Hz}$.

(c)

To determine

The wavelength of the photon.

Answer to Problem 41AAP

The wavelength of the photon is $1876\text{nm}$.

Explanation of Solution

Write the expression to calculate wavelength of the photon $\left(\lambda \right)$.

 $\lambda =\frac{hc}{E}$                                                                                                                (III)

Here, speed of light is c.

Conclusion:

The value of speed of light is $3\times {10}^8\text{m}/\text{s}$.

Substitute $1.06\times {10}^{-19}\text{J}$ for $E$, $6.63\times {10}^{-34}\text{J}\cdot \text{s}$ for $h$ and $3\times {10}^8\text{m}/\text{s}$ for c in Equation (III).

 $\begin{array}{c}\lambda =\frac{\left(6.63\times {10}^{-34}\text{J}\cdot \text{s}\right)\left(3\times {10}^8\text{m}/\text{s}\right)}{1.06\times {10}^{-19}\text{J}}\\ =\text{1}\text{.876}\times {\text{10}}^{-6}\text{m}\\ =\text{1876}\times {\text{10}}^{-9}\text{m}\\ =1876\text{nm}\end{array}$

Thus, the wavelength of the photon is $1876\text{nm}$.