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Question
The terminology of different parts of the electromagnetic spectrum is given in the text. Use the formula E=hv (for energy of a quantum of radiation photon) and obtain the photon energy in units of eV for different parts of the electromagnetic spectrum. In what ways are the different scales of photon energies that you obtain related to the sources of electromagnetic radiation.?
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Solution
Energy of photon, E=hv
This implies, E=hcλ
Where,h=6.62×10−34jsc=3×108ms−1
If wave length λ is in meter and energy is in joule then, we will divide E by 1.6×10−19 to convert into eV (Electron volt).
∴E=hcλ×1.6×10−19eV
(1) For y−rays wave length ranges from 10−10m to less that 10−14m
Therefore,
Energy=6.62×10−34×3×10810−10×1.6×10−19eV=12.4×103eV≈104eV
Thus,λ=10−10m,energy=104eV and λ=10−14m,energy=108eV
Energy of y−rays ranges between 104 to 108 eV
(2) For X−rays wave length ranges from 10−8m to 10−7mFor λ=10−8
Therefore,
Energy=6.62×10−34×3×10810−8×1.6×10−19eV=12.4≈102eV
λ=10−13m,energy=107eV
(3) For violet radiation λ ranges from 4×10−7 to 6×10−10
Therefore, for λ=4×10−7
Energy=6.62×10−34×3×1084×10−7×1.6×10−19eV=3.1eV≈1010eV
λ=6×10−10m , Energy =103eV
Energy of ultraviolet radiation vary between 1010to103eV.
(4) For visible radiations wave length range from 4×10−7mto7×10−7m
Therefore,
For λ=4×10−7m, and Energy =1010eV
Energy=6.62×10−34×3×1087×10−7×1.6×10−19eV=1.77eV≈100eV
(5) For infrared radiation λ range from 7×10−7mto7×10−14m
Therefore,
λ=7×10−7,energy=100eV
Forλ=7×10−4,energy=11000timesthe other order of 10−3eV
(6) For micro waves λ ranges from 1mm to 0.3mFor λ=1mmor10−3energy is equal to
Energy=6.62×10−34×3×10810−3×1.6×10−19eV=1.24×10−3eV≈10−3eV
Forλ=0.3m,Energy=4.1×10−6eV≈10−6eV.
(7) For Radio waves λ ranges from 1m to few kmFor λ=1m
Energy is equal to=6.62×10−34×3×108100×1.6×10−19eV=1.24×10−6eV≈10−6eV
Energy for λ of the order of few km≈10−6eV
The Energy of a photon that a source produces indicates the spacing of relevant energy levels of the source
This implies, E=hcλ
Where,
h=6.62×10−34jsc=3×108ms−1
If wave length λ is in meter and energy is in joule then, we will divide E by 1.6×10−19 to convert into eV (Electron volt).
∴E=hcλ×1.6×10−19eV
(1) For y−rays wave length ranges from 10−10m to less that 10−14m
Therefore,
Energy=6.62×10−34×3×10810−10×1.6×10−19eV=12.4×103eV≈104eV
Thus,
λ=10−10m,energy=104eV and
λ=10−14m,energy=108eV
Energy of y−rays ranges between 104 to 108 eV
(2) For X−rays wave length ranges from 10−8m to 10−7m
For λ=10−8
Therefore,
Energy=6.62×10−34×3×10810−8×1.6×10−19eV=12.4≈102eV
λ=10−13m,energy=107eV
(3) For violet radiation λ ranges from 4×10−7 to 6×10−10
Therefore,
for λ=4×10−7
Energy=6.62×10−34×3×1084×10−7×1.6×10−19eV=3.1eV≈1010eV
λ=6×10−10m , Energy =103eV
Energy of ultraviolet radiation vary between 1010to103eV.
(4) For visible radiations wave length range from 4×10−7mto7×10−7m
Therefore,
For λ=4×10−7m, and Energy =1010eV
Energy=6.62×10−34×3×1087×10−7×1.6×10−19eV=1.77eV≈100eV
(5) For infrared radiation λ range from 7×10−7mto7×10−14m
Therefore,
λ=7×10−7,energy=100eV
For
λ=7×10−4,energy=11000times
the other order of 10−3eV
(6) For micro waves λ ranges from 1mm to 0.3m
For λ=1mmor10−3
energy is equal to
Energy=6.62×10−34×3×10810−3×1.6×10−19eV=1.24×10−3eV≈10−3eV
For
λ=0.3m,Energy=4.1×10−6eV≈10−6eV.
(7) For Radio waves λ ranges from 1m to few km
For λ=1m
Energy is equal to
=6.62×10−34×3×108100×1.6×10−19eV=1.24×10−6eV≈10−6eV
Energy for λ of the order of few km≈10−6eV
The Energy of a photon that a source produces indicates the spacing of relevant energy levels of the source
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