Chapter 6, Problem 17QP

A photon has a frequency of $6.5\times {10}^9\text{Hz}$. (a) Convert this frequency into wavelength (nm). Does this frequency fall in the visible region? (b) Calculate the energy (in joules) of this photon. (c) Calculate the energy (in joules) of 1 mole of photons all with this frequency.

Interpretation Introduction

Interpretation:

The frequency of a photon is to be converted into wavelength (in nm) and the statement, “does this frequency fall in the visible region?” is to be answered with the energy of photon in joules. Also, the energy for one mole of photons having this frequency is to be evaluated.

Concept introduction:

Electromagnetic wave has an electric field component and a magnetic field component having same wavelength and frequency.

The range of wavelengths (in nm) and the type of radiation are ${\text{10}}^{-3}$ to ${10}^{-1}$ for gamma rays; ${10}^{-1}$ to $\text{10}$ for X-rays; $10$ to ${\text{10}}^3$ for ultraviolet rays; $400\text{ nm}$ to $700\text{ nm}$ for visible light; ${\text{10}}^3$ to ${\text{10}}^5$ for infrared rays; ${\text{10}}^5$ to ${\text{10}}^7$ for microwave and ${\text{10}}^7$ to ${\text{10}}^{13}$ for radio waves.

The relationship between wavelength and frequency of an electromagnetic radiation is as follows:

$\text{c}=\lambda \times \nu$

Here, $\text{c}$ is the speed of light $\left(3.00\times {10}^8\text{ m/s}\right)$, $\lambda$ is the wavelength, and $\nu$ is the frequency.

A small discrete packet of energy that is associated with electromagnetic radiation is called a photon. One photon is equivalent to one quantum.

The energy of a photon can be expressed as follows:

$\begin{array}{c}\text{E}=\text{h}\nu \\ \text{or, E}=\frac{\text{hc}}{\lambda }\end{array}$

Here, E is the energy of photon, $\text{h}$ is Planck’s constant $\left(6.63\times {10}^{-34}\text{ J}\cdot \text{s}\right)$, $\text{c}$ is the speed of light $\left(3.0\times {10}^8\text{ m/s}\right)$, $\lambda$ is the wavelength, and $\nu$ is the frequency.

The relationship between meters $\text{(m)}$ and nanometres $\text{(nm)}$ can be expressed as:

$\text{1nm = 1}\times {\text{10}}^{-9}\text{m}$

To convert meters $\text{(m)}$ to nanometres $\text{(nm)}$, conversion factor is $\frac{1\text{ nm}}{1.00\times {10}^{-9}\text{ m}}$.

To convert ${\text{(s}}^{-1}\text{)}$ to hertz $\text{(Hz)}$, conversion factor is $\frac{1.0\text{ Hz}}{1.0{\text{ s}}^{-1}}$.

The relationship between mole and number of atoms or photons can be expressed as:

$1\text{mol}\text{=}6.022\times {\text{10}}^{23}\text{photons}$.

To convert number of photons to mole, conversion factor is $\frac{1.00\text{ mol}}{6.022\times {10}^{23}\text{ photons }}$.

Answer to Problem 17QP

Solution:

(a)

$4.6\times {10}^7\text{ nm}$, the radiation does not fall in the visible region. It is in the microwave region.

(b)

$4.3\times {10}^{-24}\text{ J/photon}$

(c)

$2.6\text{ J/mol}$

Explanation of Solution

a) Convert frequency into wavelength, does it falls in visible region?

The frequency of a photon is $6.5\times {10}^9\text{ Hz}$.

The wavelength (in nm) of photon can be evaluated as follows:

$\lambda =\frac{\text{c}}{\nu }$

Substitute $3.0\times {10}^8\text{ m/s}$ for $\text{c}$ and $6.5\times {10}^9\text{ Hz}$ for $\nu$.

$\begin{array}{c}\lambda =\left(\frac{3.0\times {10}^8\cancel{\text{m}}\text{/}\cancel{\text{s}}}{6.5\times {10}^9\cancel{\text{Hz}}}\right)\left(\frac{1.0\cancel{\text{Hz}}}{1.0\cancel{{\text{s}}^{-1}}}\right)\left(\frac{1.0\text{ nm}}{1.0\times {10}^{-9}\cancel{\text{m}}}\right)\\ =4.6\times {10}^7\text{ nm}\end{array}$

Therefore, the wavelength of a photon is $4.6\times {10}^7\text{ nm}$ .

As the microwave region has a range of ${\text{10}}^5\text{ nm}$ to ${\text{10}}^7\text{nm}$, the radiation does not fall in the visible region. It falls in microwave region.

b) Energy (in joules) of photon.

The frequency of a photon is $6.5\times {10}^9\text{ Hz}$.

The energy of photon is evaluated as shown below.

$\text{E}=h\nu$

Substitute $3.0\times {10}^8\text{ m/s}$ for $\text{h}$ and $6.5\times {10}^9\text{ Hz}$ for $\nu$.

$\begin{array}{c}\text{E}=\left(6.63\times {10}^{-34}\text{ J}\cdot \cancel{\text{s}}\right)\left(6.5\times {10}^9\cancel{\text{Hz}}\right)\left(\frac{1.0\cancel{{\text{s}}^{-1}}}{1.0\cancel{\text{Hz}}}\right)\\ =4.3\times {10}^{-24}\text{ J/photon}\end{array}$

Hence, the energy of a photon is $4.3\times {10}^{-24}\text{ J/photon}$ .

c) Energy (in joules) of 1 mole of photon all with this frequency.

The frequency of a photon is $6.5\times {10}^9\text{ Hz}$.

The energy of a single photon ${\text{E}}_p$ is $4.31\times {10}^{-24}\text{ J}$.

The energy for 1 mole of photons is evaluated as shown below.

$\begin{array}{c}\text{E}=\left({\text{E}}_p\right)\left(1\text{ mol photon}\right)\\ =\left(\frac{4.31\times {10}^{-24}\text{ J}}{\text{1 }\cancel{\text{photon}}}\right)\left(\frac{6.022\times {10}^{23}\cancel{\text{photons}}}{1.00\text{ mol }}\right)\\ =2.6\text{ J/mol}\end{array}$

Hence, the energy for one mole of photons is $2.6\text{ J/mol}$ .