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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

Electrons accelerated to an energy of 50 GeVhave a de Broglie wavelength λsmall enough for them to probe the structure within a target nucleus by scattering from the structure. Assume that the energy is so large that the extreme relativistic relation p=Ecbetween momentum magnitude pand energy Eapplies. (In this extreme situation, the kinetic energy of an electron is much greater than its rest energy.)

(a) What is λ?

(b) If the target nucleus has radius R=5.0 fm, what is the ratio Rλ?

Question: You will find in Chapter 39 that electrons cannot move in definite orbits within atoms, like the planets in our solar system. To see why, let us try to “observe” such an orbiting electron by using a light microscope to measure the electron’s presumed orbital position with a precision of, say, 10pm(a typical atom has a radius of about localid="1663132292844" 100pm). The wavelength of the light used in the microscope must then be about 10pm. (a) What would be the photon energy of this light? (b) How much energy would such a photon impart to an electron in a head-on collision? (c) What do these results tell you about the possibility of “viewing” an atomic electron at two or more points along its presumed orbital path? (Hint:The outer electrons of atomsare bound to the atom by energies of only a few electron-volts.)

The existence of the atomic nucleus was discovered in 1911by Ernest Rutherford, who properly interpreted some experiments in which a beam of alpha particles was scattered from a metal foil of atoms such as gold. (a) If the alpha particles had a kinetic energy of 7.5MeV, what wads their de Broglie wavelength? (b) Explain whether the wave nature of the incident alpha particles should have been taken into account in interpreting these experiments. The mass of an alpha particle is4.00u (atomic mass units), and its distance of closest approach to the nuclear center in these experiments was about 30fm. (The wave nature of matter was not postulated until more than a decade after these crucial experiments were performed.)

Assuming that your surface temperature is98.6° Fand that you are an ideal blackbody radiator (you are close), find

(a) the wavelength at which your spectral radiancy is maximum,

(b) the power at which you emit thermal radiation in a wavelength range of 1.00 nmat that wavelength, from a surface area of4.00 cm2, and

(c) the corresponding rate at which you emit photons from that area. Using a wavelength of500 nm (in the visible range),

(d) recalculate the power and

(e) the rate of photon emission. (As you have noticed, you do not visibly glow in the dark.)

In an old-fashioned television set, electrons are accelerated through a potential difference of 25.0kV. What is the de Broglie wavelength of such electrons? (Relativity is not needed.)
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