Chapter 28, Problem 19P

(a)

To determine

The energy of the longest wavelength.

Answer to Problem 19P

The energy of the longest wavelength is $1.889\text{eV}$.

Explanation of Solution

Formula to calculate the energy of the photon of the longest wavelength is,

$E_{photon}=E_3-E_2$

• $E_2\text{and}{E}_3$ are the energy of the second and third level
• $E_{photon}$ is energy of the photon,

Substitute $\left(-1.512\text{eV}\right)$ for $E_3$, $\left(-3.401\text{eV}\right)$ for $E_2$ to find $E_{photon}$.

$\begin{array}{c}{E}_{photon}=\left(-1.512\text{eV}\right)-\left(-3.401\text{eV}\right)\\ =1.889\text{eV}\end{array}$

Thus, the energy of the photon of the longest wavelength is $1.889\text{eV}$.

Conclusion:

Therefore, the energy of the photon of the longest wavelength is $1.889\text{eV}$.

(b)

To determine

The wavelength of the photon of the longest wavelength.

Answer to Problem 19P

The wavelength of the longest wavelength is $6.58\times {10}^{-7}\text{m}$.

Explanation of Solution

Formula to calculate the wavelength of the photon of the longest wavelength is,

$\lambda =\frac{hc}{E_{photon}}$

• $h$ is the Planck’s constant
• $c$ is the speed of light
• $E_{photon}$ is energy of the photon,

From unit conversion,

$1\text{eV}=1.6\times {10}^{-19}\text{J}$

Substitute $6.63\times {10}^{-34}\text{J-s}$ for $h$, $3\times {10}^8\text{m/s}$ for $c$, $1.889\text{eV}$ for $E_{photon}$ to find $\lambda$.

$\begin{array}{c}\lambda =\frac{\left(6.63\times {10}^{-34}\text{J-s}\right)\left(3\times {10}^8\text{m/s}\right)}{\left(1.889\text{eV}\right)}\left(\frac{1\text{eV}}{1.6\times {10}^{-19}\text{J}}\right)\\ =6.58\times {10}^{-7}\text{m}\end{array}$

Thus, the wavelength of the photon of the longest wavelength is $6.58\times {10}^{-7}\text{m}$.

Conclusion:

Therefore, the wavelength of the photon of the longest wavelength is $6.58\times {10}^{-7}\text{m}$.

(c)

To determine

The energy of the shortest photon wavelength.

Answer to Problem 19P

The energy of the shortest photon wavelength is $3.02\text{eV}$.

Explanation of Solution

Formula to calculate the energy of the photon of the shortest wavelength is,

$E_{photon}=E_6-E_2$

• $E_2\text{and}{E}_6$ are the energy of the second and sixth level
• $E_{photon}$ is energy of the photon,

Substitute $\left(-0.378\text{eV}\right)$ for $E_6$, $\left(-3.401\text{eV}\right)$ for $E_2$ to find $E_{photon}$.

$\begin{array}{c}{E}_{photon}=\left(-0.378\text{eV}\right)-\left(-3.401\text{eV}\right)\\ =3.02\text{eV}\end{array}$

Thus, the energy of the photon of the shortest wavelength is $3.02\text{eV}$.

Conclusion:

Therefore, the energy of the photon of the shortest wavelength is $3.02\text{eV}$.

(d)

To determine

The wavelength of the photon of the shortest wavelength.

Answer to Problem 19P

The wavelength of the shortest wavelength is $4.12\times {10}^{-7}\text{m}$.

Explanation of Solution

Formula to calculate the wavelength of the photon of the shortest wavelength is,

$\lambda =\frac{hc}{E_{photon}}$

• $h$ is the Planck’s constant
• $c$ is the speed of light
• $E_{photon}$ is energy of the photon,

From unit conversion,

$1\text{eV}=1.6\times {10}^{-19}\text{J}$

Substitute $6.63\times {10}^{-34}\text{J-s}$ for $h$, $3\times {10}^8\text{m/s}$ for $c$, $3.02\text{eV}$ for $E_{photon}$ to find $\lambda$.

$\begin{array}{c}\lambda =\frac{\left(6.63\times {10}^{-34}\text{J-s}\right)\left(3\times {10}^8\text{m/s}\right)}{\left(3.02\text{eV}\right)}\left(\frac{1\text{eV}}{1.6\times {10}^{-19}\text{J}}\right)\\ =4.12\times {10}^{-7}\text{m}\end{array}$

Thus, the wavelength of the photon of the shortest wavelength is $4.12\times {10}^{-7}\text{m}$.

Conclusion:

Therefore, the wavelength of the photon of the shortest wavelength is $4.12\times {10}^{-7}\text{m}$.

(e)

To determine

The shortest possible wavelength in the Balmar series.

Answer to Problem 19P

The shortest possible wavelength in the Balmar series is $3.66\times {10}^{-7}\text{m}$.

Explanation of Solution

Section1:

To determine: The energy of the shortest possible wavelength.

Answer: The energy of the shortest possible wavelength. is $3.401\text{eV}$.

Explanation:

Formula to calculate the energy of the photon of the shortest wavelength is,

$E_{photon}=E_{\infty }-E_2$

• $E_2\text{and}{E}_{\infty }$ are the energy of the second and infinite level
• $E_{photon}$ is energy of the photon,

Substitute $\left(0\text{eV}\right)$ for $E_{\infty }$, $\left(-3.401\text{eV}\right)$ for $E_2$ to find $E_{photon}$.

$\begin{array}{c}{E}_{photon}=\left(\text{0}\text{eV}\right)-\left(-3.401\text{eV}\right)\\ =3.401\text{eV}\end{array}$

Thus, the energy of the photon of the shortest wavelength is $3.401\text{eV}$.

Section 2:

To determine: The shortest wavelength.

Answer: The shortest wavelength is $3.66\times {10}^{-7}\text{m}$.

Explanation:

Formula to calculate the wavelength of the photon of the shortest wavelength is,

$\lambda =\frac{hc}{E_{photon}}$

• $h$ is the Planck’s constant
• $c$ is the speed of light
• $E_{photon}$ is energy of the photon,

From unit conversion,

$1\text{eV}=1.6\times {10}^{-19}\text{J}$

Substitute $6.63\times {10}^{-34}\text{J-s}$ for $h$, $3\times {10}^8\text{m/s}$ for $c$, $3.401\text{eV}$ for $E_{photon}$ to find $\lambda$.

$\begin{array}{c}\lambda =\frac{\left(6.63\times {10}^{-34}\text{J-s}\right)\left(3\times {10}^8\text{m/s}\right)}{\left(3.4012\text{eV}\right)}\left(\frac{1\text{eV}}{1.6\times {10}^{-19}\text{J}}\right)\\ =3.66\times {10}^{-7}\text{m}\end{array}$

Thus, the shortest wavelength is $3.66\times {10}^{-7}\text{m}$.

Conclusion:

Therefore, the shortest wavelength is $3.66\times {10}^{-7}\text{m}$