If a neon atom has an transition that emits light with a wavelength of $7.33xx10^7$ m, what color of light is given off? How much energy is given off?

Md Ariful Islam

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Md Ariful Islam

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$7.33xx10^7"m"$ is not a wavelength in the visible light electromagnetic spectrum. It would be a radio wave because they have the longest wavelengths.

Calculate energy.

We can still calculate energy using the equation:

$E=(hc)/lambda$

where $E$ is energy in Joules (J), $c$ is the speed of light, and

$lambda$ (pronounced "lambda") is the wavelength.

$h= 6.62606896xx10^(−34) "J·s"$, and is known as .

$c="299792458 m/s"$, and is the speed of light in a vacuum.

Substitute the known values into the equation.

$E=(6.62606896xx10^(−34) "J"·cancel"s"xx299792458 cancel("m")/cancel"s")/(7.33xx10^7cancel"m")=2.71xx10^(-33) "J"$ (rounded to three significant figures)

Many people use $3.00xx10^8$ m/s for the speed of light rounded to three , and $6.626xx10^(-34)$ $"J"*"s"$ for Planck's constant rounded to four significant figures.

The wavelength $7.33xx10^7$ m is in the Extremely Low Frequency (ELF) end of the electromagnetic spectrum, which are radio waves Megameters (Mm) in length.

$"1 Mm"=1xx10^6" m"$

$(7.33xx10^7cancel"m")xx(1"Mm")/(1xx10^6cancel"m")="73.3 Mm"$

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