Answer to Question #109608 in Inorganic Chemistry for Delaney Gilder

Question #109608
Determine the energy of a photon with of 361 nm? (h= 6.626 x 10^-34 J x s and c= 3.00 x 10^8 m/s)
1
Expert's answer
2020-04-15T01:51:26-0400

The energy E of a photon of a frequency v is E=hxv, where h is Plank's constant (6.626x10^-34 Jxs). Frequency v is related to wavelength lambda (361nm or 3.61x10^-7 m) through speed of light c (3x10^8 m/s) as v=c/lambda. Substituting this into formula for energy yields E=hxc/lambda, then

E = (6.626x10^-34 Jxs) x (3x10^8 m/s) / (3.61x10^-7 m) = 5.5x10^-19 J

Answer is 5.5 x 10^-19 J


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