 # The blue color of the sky results from the scattering Sam Longoria 2021-12-15 Answered
The blue color of the sky results from the scattering of sunlight by molecules in the air. The blue light has a frequency of about $7.5×{10}^{14}Hz$
a) Calculate the wavelength (in nm) associated with this radiation, and
b) calculate the energy (in joules) of a single photon associated with this frequency.
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Step 1
Part A:
Use the speed of light equation to calculate the wavelength corresponding to this frequency.
$x=\lambda \nu$
$⇒\lambda =\frac{c}{\nu }$
$=\frac{2.998×{10}^{8}\frac{m}{s}}{7.5×{10}^{14}{s}^{-1}}$
$=4.0×{10}^{-7}m$
Determine the wavelength of this photon in units of nanometers:
1. Begin with the wavelength in meters.
2. Use a conversion factor to convert meters into nanometers.

$=4.0×{10}^{2}nm$
Step 2
Part B:
Use the equation for the energy of a photon to calculate the energy of this photon
$E=h\nu$
$=\left(6.626×{10}^{-34}J×s\right)\left(7.5×{10}^{14}{s}^{-1}\right)$
$=5.0×{10}^{-19}J$

###### Not exactly what you’re looking for? Anzante2m
Step 1
Given:
The frequency of light $=7.5×{10}^{14}Hz$
Calculating the wavelength:
Relation between frequency and wavelength: $\lambda =\frac{\nu }{c}$
$c=3×{10}^{8}\frac{m}{s}$
$\nu =7.5×{10}^{14}Hz$
Substitute the known values in the above formula and simplify,
$\lambda =\frac{7.5×{10}^{14}Hz}{3×{10}^{8}\frac{m}{s}}$
$=0.4×{10}^{-6}m$
$=400×{10}^{-9}m$
And, $1nm={10}^{-9}m$
$=400nm$
$\lambda =400nm$
Step 2
The energy of a single photon associated with the frequency:
Formula: $E=h\nu$
$h=\text{placonstant}=6.63×{10}^{-34}$
$\nu =7.5×{10}^{14}Hz$
$E=6.63×{10}^{-34}×7.5×{10}^{14}Hz$
$=49.695×{10}^{-20}J$
$=4.9695×{10}^{-19}J$
$E=4.9695×{10}^{-19}J$