# 5.3 Photon energies and the electromagnetic spectrum  (Page 6/15)

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## Phet explorations: color vision

Make a whole rainbow by mixing red, green, and blue light. Change the wavelength of a monochromatic beam or filter white light. View the light as a solid beam, or see the individual photons.

## Section summary

• Photon energy is responsible for many characteristics of EM radiation, being particularly noticeable at high frequencies.
• Photons have both wave and particle characteristics.

## Conceptual questions

Why are UV, x rays, and $\gamma$ rays called ionizing radiation?

How can treating food with ionizing radiation help keep it from spoiling? UV is not very penetrating. What else could be used?

Some television tubes are CRTs. They use an approximately 30-kV accelerating potential to send electrons to the screen, where the electrons stimulate phosphors to emit the light that forms the pictures we watch. Would you expect x rays also to be created?

Tanning salons use “safe” UV with a longer wavelength than some of the UV in sunlight. This “safe” UV has enough photon energy to trigger the tanning mechanism. Is it likely to be able to cause cell damage and induce cancer with prolonged exposure?

Your pupils dilate when visible light intensity is reduced. Does wearing sunglasses that lack UV blockers increase or decrease the UV hazard to your eyes? Explain.

One could feel heat transfer in the form of infrared radiation from a large nuclear bomb detonated in the atmosphere 75 km from you. However, none of the profusely emitted x rays or $\gamma$ rays reaches you. Explain.

Can a single microwave photon cause cell damage? Explain.

In an x-ray tube, the maximum photon energy is given by $\text{hf}=\text{qV}.$ Would it be technically more correct to say $\text{hf}=\text{qV}+\text{BE,}$ where BE is the binding energy of electrons in the target anode? Why isn’t the energy stated the latter way?

## Problems&Exercises

What is the energy in joules and eV of a photon in a radio wave from an AM station that has a 1530-kHz broadcast frequency?

$6.34×{\text{10}}^{-9}\phantom{\rule{0.25em}{0ex}}\text{eV}$ , $1.01×{\text{10}}^{-27}\phantom{\rule{0.25em}{0ex}}\text{J}$

(a) Find the energy in joules and eV of photons in radio waves from an FM station that has a 90.0-MHz broadcast frequency. (b) What does this imply about the number of photons per second that the radio station must broadcast?

Calculate the frequency in hertz of a 1.00-MeV $\gamma$ -ray photon.

$2\text{.}\text{42}×{\text{10}}^{\text{20}}\phantom{\rule{0.25em}{0ex}}\text{Hz}$

(a) What is the wavelength of a 1.00-eV photon? (b) Find its frequency in hertz. (c) Identify the type of EM radiation.

Do the unit conversions necessary to show that $\text{hc}=\text{1240 eV}\cdot \text{nm,}$ as stated in the text.

$\begin{array}{lll}\text{hc}& =& \left(\text{6.62607}×{\text{10}}^{-\text{34}}\phantom{\rule{0.25em}{0ex}}J\cdot s\right)\left(\text{2.99792}×{\text{10}}^{8}\phantom{\rule{0.25em}{0ex}}\text{m/s}\right)\left(\frac{{\text{10}}^{9}\phantom{\rule{0.25em}{0ex}}\text{nm}}{1 m}\right)\left(\frac{\text{1.00000 eV}}{\text{1.60218}×{\text{10}}^{-\text{19}}\phantom{\rule{0.25em}{0ex}}\text{J}}\right)\\ & =& \text{1239.84 eV}\cdot \text{nm}\\ & \approx & \text{1240 eV}\cdot \text{nm}\end{array}$

Confirm the statement in the text that the range of photon energies for visible light is 1.63 to 3.26 eV, given that the range of visible wavelengths is 380 to 760 nm.

(a) Calculate the energy in eV of an IR photon of frequency $\text{2.00}×{\text{10}}^{\text{13}}\phantom{\rule{0.25em}{0ex}}\text{Hz.}$ (b) How many of these photons would need to be absorbed simultaneously by a tightly bound molecule to break it apart? (c) What is the energy in eV of a $\gamma$ ray of frequency $3\text{.}\text{00}×{\text{10}}^{\text{20}}\phantom{\rule{0.25em}{0ex}}\text{Hz?}$ (d) How many tightly bound molecules could a single such $\gamma$ ray break apart?

(a) 0.0829 eV

(b) 121

(c) 1.24 MeV

(d) $1\text{.}\text{24}×{\text{10}}^{5}$

Prove that, to three-digit accuracy, $h=4\text{.}\text{14}×{\text{10}}^{-\text{15}}\phantom{\rule{0.25em}{0ex}}\text{eV}\cdot s,$ as stated in the text.

(a) What is the maximum energy in eV of photons produced in a CRT using a 25.0-kV accelerating potential, such as a color TV? (b) What is their frequency?

(a) $\text{25.0}×{\text{10}}^{3}\phantom{\rule{0.25em}{0ex}}\text{eV}$

(b) $\text{6}\text{.}\text{04}×{\text{10}}^{\text{18}}\phantom{\rule{0.25em}{0ex}}\text{Hz}$

What is the accelerating voltage of an x-ray tube that produces x rays with a shortest wavelength of 0.0103 nm?

(a) What is the ratio of power outputs by two microwave ovens having frequencies of 950 and 2560 MHz, if they emit the same number of photons per second? (b) What is the ratio of photons per second if they have the same power output?

(a) 2.69

(b) 0.371

How many photons per second are emitted by the antenna of a microwave oven, if its power output is 1.00 kW at a frequency of 2560 MHz?

Some satellites use nuclear power. (a) If such a satellite emits a 1.00-W flux of $\gamma$ rays having an average energy of 0.500 MeV, how many are emitted per second? (b) These $\gamma$ rays affect other satellites. How far away must another satellite be to only receive one $\gamma$ ray per second per square meter?

(a) $\text{1}\text{.}\text{25}×{\text{10}}^{\text{13}}\phantom{\rule{0.25em}{0ex}}\text{photons/s}$

(b) 997 km

(a) If the power output of a 650-kHz radio station is 50.0 kW, how many photons per second are produced? (b) If the radio waves are broadcast uniformly in all directions, find the number of photons per second per square meter at a distance of 100 km. Assume no reflection from the ground or absorption by the air.

How many x-ray photons per second are created by an x-ray tube that produces a flux of x rays having a power of 1.00 W? Assume the average energy per photon is 75.0 keV.

$\text{8.33}×{\text{10}}^{\text{13}}\phantom{\rule{0.25em}{0ex}}\text{photons/s}$

(a) How far away must you be from a 650-kHz radio station with power 50.0 kW for there to be only one photon per second per square meter? Assume no reflections or absorption, as if you were in deep outer space. (b) Discuss the implications for detecting intelligent life in other solar systems by detecting their radio broadcasts.

Assuming that 10.0% of a 100-W light bulb’s energy output is in the visible range (typical for incandescent bulbs) with an average wavelength of 580 nm, and that the photons spread out uniformly and are not absorbed by the atmosphere, how far away would you be if 500 photons per second enter the 3.00-mm diameter pupil of your eye? (This number easily stimulates the retina.)

181 km

Construct Your Own Problem

Consider a laser pen. Construct a problem in which you calculate the number of photons per second emitted by the pen. Among the things to be considered are the laser pen’s wavelength and power output. Your instructor may also wish for you to determine the minimum diffraction spreading in the beam and the number of photons per square centimeter the pen can project at some large distance. In this latter case, you will also need to consider the output size of the laser beam, the distance to the object being illuminated, and any absorption or scattering along the way.

#### Questions & Answers

so some one know about replacing silicon atom with phosphorous in semiconductors device?
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SUYASH
What is lattice structure?
of graphene you mean?
Ebrahim
or in general
Ebrahim
in general
s.
Graphene has a hexagonal structure
tahir
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Cied
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China
Cied
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Porter
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Porter
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Yasmin
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Cesar
I'm interested in nanotube
Uday
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what is system testing
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Stotaw
In this morden time nanotechnology used in many field . 1-Electronics-manufacturad IC ,RAM,MRAM,solar panel etc 2-Helth and Medical-Nanomedicine,Drug Dilivery for cancer treatment etc 3- Atomobile -MEMS, Coating on car etc. and may other field for details you can check at Google
Azam
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Prasenjit
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Azam
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Prasenjit
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Damian
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Damian
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Azam
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Uday
I'm interested in Nanotube
Uday
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Prasenjit
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At high concentrations (>0.01 M), the relation between absorptivity coefficient and absorbance is no longer linear. This is due to the electrostatic interactions between the quantum dots in close proximity. If the concentration of the solution is high, another effect that is seen is the scattering of light from the large number of quantum dots. This assumption only works at low concentrations of the analyte. Presence of stray light.
the Beer law works very well for dilute solutions but fails for very high concentrations. why?
how did you get the value of 2000N.What calculations are needed to arrive at it
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