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Chapter 38: Q 80P (page 1185)

Question: A spectral emission line is electromagnetic radiation that is emitted

in a wavelength range narrow enough to be taken as a single wavelength. One such emission line that is important in astronomyhas a wavelength of 21cm . What is the photon energy in the electromagnetic wave at that wavelength?

Short Answer

Expert verified

The photon energy in the electromagnetic wave is5.9μeV.

Step by step solution

01

Identifying the data given in the question

The wavelength of the electromagnetic wave is λ=21cm

02

Concept used to solve the question

Photon Energy:

The energy that a single photon carries is known as photon energy. Energy is inversely correlated with wavelength because it is directly proportional to the electromagnetic frequency of the photon

03

Finding the photon energy

The photon energy in the electromagnetic wave can be given as

E=hcλ

Where his plank constant,c is the velocity of light, and λis the wavelength

We know role="math" localid="1663228885797" hc=1240nm·eV

Substituting the values

E=1240nm·eV21×107nm=5.9×10-6eV=5.9μeV

Hence the photon energy in the electromagnetic wave is5.9μeV.

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Most popular questions from this chapter

Question: Figure 38-13 shows a case in which the momentum component

pxof a particle is fixed so that ∆px=0 ; then, from Heisenberg’s uncertainty principle (Eq. 38-28), the position x of the particle is completely unknown. From the same principle it follows that the opposite is also true; that is, if the position of a particle is exactly known (∆x=0), the uncertainty in its momentum is infinite.Consider an intermediate case, in which the position of aparticle is measured, not to infinite precision, but to within a distanceof λ2π, where λis the particle’s de Broglie wavelength.Show that the uncertainty in the (simultaneously measured) momentumcomponent is then equal to the component itself; that is,∆px=p. Under these circumstances, would a measured momentumof zero surprise you? What about a measured momentum of 0.5p? Of 0.2p? Of 12p?

Show that|ψ|2=|ψ|2 , with related as in Eq. 38-14. That is, show that the probability density does not depend on the time variable.

A light detector has an absorbing area of 2.00×10-6m2 and absorbs 50% of the incident light, which is at a wavelength 600nm. The detector faces an isotropic source, 12.0 m from the source. The energy E emitted by the source versus time t is given in Fig. 38-26 ( Es=7.2nJ, ts=2.0s ). At what rate are photons absorbed by the detector?

The smallest amount of energy needed to eject an electron from metallic sodium is 2.28eV. Does sodium show a photoelectric effect for red light, with λ=680nm? (That is, does the light cause electron emission?) (b) What is the cutoff wavelength for photoelectric emission from sodium? (c) To what color does that wavelength correspond?

Question:Consider a potential energy barrier like that of Fig. 38-17but whose height Ubis and 6eVwhose thickness Lis 0.70nm. What is the energy of an incident electron whose transmissioncoefficient is 0.0010?
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