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Chapter 8: Problem 21

Without doing detailed calculations, indicate which of the following electromagnetic radiations has the greatest energy per photon and which has the least: (a) \(662 \mathrm{nm}\) (b) \(2.1 \times 10^{-5} \mathrm{cm} ;\) (c) \(3.58 \mu \mathrm{m} ;\) (d) \(4.1 \times 10^{-6} \mathrm{m}\).

Short Answer

Expert verified
Photon (d) with wavelength \(4.1 \times 10^{-6} m\) has the greatest energy and photon (b) with wavelength \(2.1 \times 10^{-7} m\) has the least energy.

Step by step solution

01

Transform Wavelength Units

First, convert all wavelengths to the same unit for a proper comparison. The comprehensive unit will be meter (m) - the SI unit for length. The conversions are: (a) \(662 nm = 662 \times 10^{-9} m\), (b) \(2.1 \times 10^{-5} cm = 2.1 \times 10^{-7} m\), (c) \(3.58 \mu m = 3.58 \times 10^{-6} m\), (d) \(4.1 \times 10^{-6} m\) remains the same.
02

Identify the Photon with Greatest Energy

To identify the photon with the greatest energy, look for the radiation with the shortest wavelength. As calculated in the previous step, (d) \(4.1 \times 10^{-6} m\) is the shortest wavelength.
03

Identify the Photon with Least Energy

To identify the photon with the least energy, look for the radiation with the longest wavelength. As calculated in step 1, (b) \(2.1 \times 10^{-7} m\) is the longest wavelength.

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Most popular questions from this chapter

Show that the probability of finding a \(2 p_{y}\) electron in the \(x z\) plane is zero.

Use Planck's equation (8.3) to determine (a) the energy, in joules per photon, of radiation of frequency $7.39 \times 10^{15} \mathrm{s}^{-1}$ (b) the energy, in kilojoules per mole, of radiation of frequency $1.97 \times 10^{14} \mathrm{s}^{-1}$

Describe two ways in which the orbitals of multielectron atoms resemble hydrogen orbitals and two ways in which they differ from hydrogen orbitals.

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