A monochromatic green light source used in a diffraction experiment has a wavelength of $5.80 *10^-7$ m. What is the frequency of the light? What is the energy of a photon of this light?

Md Ariful Islam

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Md Ariful Islam

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The inverse relationship that exists between wavelength and frequency is described by the equation

$color(blue)( lamda * nu = c)" "$, where

$lamda$ - the frequency of the wave
$nu$ - its frequency
$c$ - the speed of light in vacuum, usually given as $3 * 10^8"ms"^(-1)$

Notice that the wavelength of the green light if given in meters and that the units used for the speed of light are meters per second. This tells you that you don't have to do any before plugging in your value into the above equation.

So, the frequency of this green light will be

$lamda * nu = c implies nu = c/(lamda)$

$nu = (3 * 10^8 color(red)(cancel(color(black)("m"))) "s"^(-1))/(5.80 * 10^(-7)color(red)(cancel(color(black)("m")))) = 5.17 * 10^(14)"s"^(-1)$

As you know,

$"1 Hz" = "1 s"^(-1)$

Expressed in Hertz, the frequency of the light will be

$nu = color(green)(5.17 * 10^(14)"Hz"$

Now, the energy of an electromagnetic wave is directly proportional to that wave's frequency. This relationship is described by the Planck - Einstein equation, which looks like this

$color(blue)(E = h * nu)" "$, where

$E$ - the energy of the wave
$h$ - , equal to $6.626 * 10^(-34)"J s"$
$nu$ - the frequency of the wave

Notice that uses Joules second, which means that you're going to have to go back to $"s"^(-1)$ for the units of the frequency.

Once again, plug in your value to get

$E = 6.626 * 10^(-34)"J" color(red)(cancel(color(black)("s"))) * 5.17 * 10^(14)color(red)(cancel(color(black)("s"^(-1))))$

$E = color(green)(3.43 * 10^(-19)"J")$

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