Chapter 22, Problem 5P

(a)

To determine

How many minutes does it take a photon to travel from the Sun to the Earth.

Answer to Problem 5P

The time taken by the photon to travel from the Sun to Earthis $8.33\text{min}$.

Explanation of Solution

The equation for time is,

$\Delta t=\frac{d}{c}$

• $d$ is the distance
• $c$ is the speed of light

Substitute $1.50\times {10}^{11}\text{m}$ for $d$ and $3.00\times {10}^8\text{m/s}$ for $c$.

$\begin{array}{c}\Delta t=\frac{1.50\times {10}^{11}\text{m}}{3.00\times {10}^8\text{m/s}}\\ =\left(500\text{s}\right)\left(\frac{1\text{min}}{60\text{s}}\right)\\ =\text{8}\text{.33}\text{min}\end{array}$

Conclusion:

Thus, the time taken by the photon to travel from the Sun to Earth is $\underset{\_}{8.33\text{min}}$.

(b)

To determine

The energy of the photon.

Answer to Problem 5P

The energy of the photon is $2.23\text{eV}$.

Explanation of Solution

The energy of a photon is,

$E=\frac{hc}{\lambda }$

• $h$ is the Plank’s constant
• $c$ is the speed of light
• $\lambda$ is the wavelength

Substitute $6.63\times {10}^{-34}\text{J}\text{.s}$ for $h$, $3.00\times {10}^8\text{m/s}$ for $c$ and $558\text{nm}$ for $\lambda$.

$\begin{array}{c}E=\frac{\left(6.63\times {10}^{-34}\text{J}\text{.s}\right)\left(3.00\times {10}^8\text{m/s}\right)}{\left(558\text{nm}\right)\left(\frac{{10}^{-9}\text{m}}{1\text{nm}}\right)}\left(\frac{1\text{eV}}{1.60\times {10}^{-19}\text{J}}\right)\\ =2.23\text{eV}\end{array}$

Conclusion:

Thus, the energy of the photon is $\underset{\_}{2.23\text{eV}}$.

(c)

To determine

The wavelength of the photon.

Answer to Problem 5P

The wavelength of the photon is $1.24\times {10}^{-6}\text{m}$.

Explanation of Solution

The wavelength of a photon is,

$\lambda =\frac{hc}{E}$

• $h$ is the Plank’s constant
• $c$ is the speed of light
• $E$ is the energy

Substitute $6.63\times {10}^{-34}\text{J}\text{.s}$ for $h$, $3.00\times {10}^8\text{m/s}$ for $c$ and $1.00\text{eV}$ for $\lambda$.

$\begin{array}{c}\lambda =\frac{\left(6.63\times {10}^{-34}\text{J}\text{.s}\right)\left(3.00\times {10}^8\text{m/s}\right)}{\left(1.00\text{eV}\right)\left(\frac{1.60\times {10}^{-19}\text{J}}{1\text{eV}}\right)}\\ =1.24\times {10}^{-6}\text{m}\end{array}$

Conclusion:

Thus, the wavelength of the photon is $\underset{\_}{1.24\times {10}^{-6}\text{m}}$.