Chapter 3, Problem 16Q

Interpretation Introduction

Interpretation:

Increasing order of energy for the given different wavelength has to be arranged; and the wavelength possesses the most energetic photons has to be identified.

Concept-Introduction:

Wavelength: The distance between successive peaks in a wave and measured in units of length.

Frequency: The number of waves passed a point in a certain amount of time.

Relation between wavelength and frequency:

$\text{Frequency}\left(\text{ν}\right)\text{=}\frac{\text{speed}\text{of}\text{light}\left(\text{c}\right)}{\text{wavelength}\left(\text{λ}\right)}$

Where, c is constant, the value of c is $\left(\text{3}{\text{.00×10}}^{\text{8}}\text{m/s}\right)$ and represents the maximum velocity that light is able to travel through air.

Conversion of centimeter into meter (cm to m): $\text{1 cm}\text{=}{\text{1×10}}^{\text{-2}}\text{m}$

Explanation of Solution

Option (a): The wavelength is $\text{2}\text{.0 cm}$

Conversion of centimeter to meter is,

$\begin{array}{l}\text{1 cm}\text{=}{\text{1×10}}^{\text{-2}}\text{m}\\ \\ \text{2}\text{.0 cm}\text{=}?\\ \\ =\frac{\left({\text{1×10}}^{\text{-2}}\text{m}\right)\left(2.0\text{cm}\right)}{\text{1 cm}}=2\times 10^{-2}m\end{array}$

Therefore, the given value of wavelength in meter is $2\times 10^{-2}m$ which fall in the region of Microwave region of the spectrum.

Calculating energy of the particular wavelength as follows,

$\text{E}\text{=}\text{hν}$

Where, h is Planck constant, $\text{ν}$ is frequency.

Substituting this relation $\text{Frequency}\left(\text{ν}\right)\text{=}\frac{\text{speed}\text{of}\text{light}\left(\text{c}\right)}{\text{wavelength}\left(\text{λ}\right)}$; energy becomes,

$\text{E}\text{=}\text{h}\frac{c}{\lambda }$

Where, c is constant, the value of c is $\left(\text{3}{\text{.00×10}}^{\text{8}}\text{m/s}\right)$; $\lambda$ is wavelength.

$\begin{array}{c}\text{E}\text{=}\text{h}\frac{\text{c}}{\text{λ}}\\ \\ \text{E}\text{=}\left(\text{6}\text{.626}\text{×}{\text{10}}^{\text{-34}}\text{J}\text{.sec}\right)\frac{\text{3×}{\text{10}}^{\text{8}}m.s^{-1}}{{\text{2×10}}^{\text{-2}}\text{m}}\\ \\ \text{=}\text{9}\text{.94 }\text{×}{\text{10}}^{\text{-24}}\text{J}\simeq 1.0\times 10^{-24}J\end{array}$

The energy of given wavelength is $1.0\times 10^{-24}J$

Option (b): The wavelength is $\text{50}\mu \text{m}$

Conversion of micrometer to meter is,

$\begin{array}{l}\text{1 μm}\text{=}{\text{1×10}}^{\text{-6}}\text{m}\\ \\ \text{50 }\mu \text{m}\text{=}?\\ \\ =\frac{\left({\text{1×10}}^{\text{-6}}\text{m}\right)\left(50\mu m\right)}{\text{1 }\mu \text{m}}=50\times 10^{-6}m\end{array}$

Therefore, the given value of wavelength in meter is $50\times 10^{-6}m$ which fall in the region of visible region of the spectrum.

Calculating energy of the particular wavelength as follows,

$\text{E}\text{=}\text{hν}$

Where, h is Planck constant, $\text{ν}$ is frequency.

Substituting this relation $\text{Frequency}\left(\text{ν}\right)\text{=}\frac{\text{speed}\text{of}\text{light}\left(\text{c}\right)}{\text{wavelength}\left(\text{λ}\right)}$; energy becomes,

$\text{E}\text{=}\text{h}\frac{c}{\lambda }$

Where, c is constant, the value of c is $\left(\text{3}{\text{.00×10}}^{\text{8}}\text{m/s}\right)$; $\lambda$ is wavelength.

$\begin{array}{c}\text{E}\text{=}\text{h}\frac{\text{c}}{\text{λ}}\\ \\ \text{E}\text{=}\left(\text{6}\text{.626}\text{×}{\text{10}}^{\text{-34}}\text{J}\text{.sec}\right)\frac{\text{3×}{\text{10}}^{\text{8}}m.s^{-1}}{{\text{50×10}}^{\text{-6}}\text{m}}\\ \\ \text{=}\text{0}\text{.39756}\text{×}{\text{10}}^{\text{-20}}\text{J}\simeq 4.0\times 10^{-21}J\end{array}$

The energy of given wavelength is $4.0\times 10^{-21}J$

Option (c): The given wavelength is $\text{400}\text{nm}$

Conversion of nanometer to meter is,

$\begin{array}{l}\text{1 nm}\text{=}{\text{1×10}}^{\text{-9}}\text{m}\\ \\ \text{400}\text{nm}\text{=}?\\ \\ =\frac{\left({\text{1×10}}^{\text{-9}}\text{m}\right)\left(400nm\right)}{\text{1 nm}}=40\times 10^{-8}m\end{array}$

Therefore, the given value of wavelength in meter is $40\times 10^{-8}m$ which fall in the region of violet region of visible light of the spectrum. Because visible region begins from the range $10^{-6}-10^{-8}m$, so the obtained range falls at the violet color.

Calculating energy of the particular wavelength as follows,

$\text{E}\text{=}\text{hν}$

Where, h is Planck constant, $\text{ν}$ is frequency.

Substituting this relation $\text{Frequency}\left(\text{ν}\right)\text{=}\frac{\text{speed}\text{of}\text{light}\left(\text{c}\right)}{\text{wavelength}\left(\text{λ}\right)}$; energy becomes,

$\text{E}\text{=}\text{h}\frac{c}{\lambda }$

Where, c is constant, the value of c is $\left(\text{3}{\text{.00×10}}^{\text{8}}\text{m/s}\right)$; $\lambda$ is wavelength.

$\begin{array}{c}\text{E}\text{=}\text{h}\frac{\text{c}}{\text{λ}}\\ \\ \text{E}\text{=}\left(\text{6}\text{.626}\text{×}{\text{10}}^{\text{-34}}\text{J}\text{.sec}\right)\frac{\text{3×}{\text{10}}^{\text{8}}m.s^{-1}}{{\text{40×10}}^{\text{-8}}\text{m}}\\ \\ \text{=}0.\text{496 }\text{×}{\text{10}}^{\text{-19}}\text{J}\simeq 5.0\times 10^{-19}J\end{array}$

The energy of given wavelength is $5.0\times 10^{-19}J$

Option (d): The given wavelength is $\text{150}\text{mm}$

Conversion of millimeter to meter is,

$\begin{array}{l}\text{1 mm}\text{=}{\text{1×10}}^{\text{-3}}\text{m}\\ \\ \text{150}\text{mm}\text{=}\text{?}\\ \\ \text{=}\frac{\left({\text{1×10}}^{\text{-3}}\text{m}\right)\left(\text{150 mm}\right)}{\text{1 mm}}\text{=}15\times 10^{-2}m\end{array}$

Therefore, the given value of wavelength in meter is $15\times 10^{-2}m$ which fall in the region of microwave region of the spectrum.

Calculating energy of the particular wavelength as follows,

$\text{E}\text{=}\text{hν}$

Where, h is Planck constant, $\text{ν}$ is frequency.

Substituting this relation $\text{Frequency}\left(\text{ν}\right)\text{=}\frac{\text{speed}\text{of}\text{light}\left(\text{c}\right)}{\text{wavelength}\left(\text{λ}\right)}$; energy becomes,

$\text{E}\text{=}\text{h}\frac{c}{\lambda }$

Where, c is constant, the value of c is $\left(\text{3}{\text{.00×10}}^{\text{8}}\text{m/s}\right)$; $\lambda$ is wavelength.

$\begin{array}{c}\text{E}\text{=}\text{h}\frac{\text{c}}{\text{λ}}\\ \\ \text{E}\text{=}\left(\text{6}\text{.626}\text{×}{\text{10}}^{\text{-34}}\text{J}\text{.sec}\right)\frac{\text{3×}{\text{10}}^{\text{8}}m.s^{-1}}{{\text{15×10}}^{\text{-2}}\text{m}}\\ \\ \text{=}\text{1}\text{.325 }\text{×}{\text{10}}^{\text{-24}}\text{J}\simeq 1.3\times 10^{-24}J\end{array}$

The energy of given wavelength is $1.3\times 10^{-24}J$

Hence, the energy as follows,

Energy of option (a) is $1.0\times 10^{-24}J$

Energy of option (b) is $4.0\times 10^{-21}J$

Energy of option (c) is $5.0\times 10^{-19}J$

Energy of option (d) is $1.3\times 10^{-24}J$

Therefore, the increasing order of energy of the given wavelength is,

$\text{(A)}\text{<}\text{(D)}\text{<}\text{(B)}\text{<}\text{(C)}$

The wavelength of the most energetic photons is $\text{400}\text{nm}$