What is the wavelength of light (in nm) if 1.21*10^21 photons have a total energy of 483 J?

1 Answer
Jan 31, 2018

lamda=498nm

Explanation:

First, we need to know the energy of one photon (assuming they all have the same energy).

"Energy of one photon"=483/(1.21*10^21)~~3.99*10^(-19)J

E=hf=(hc)/lamda

3.99*10^-19=((6.63*10^-34)(3*10^8))/lamda

lamda=((6.63*10^-34)(3*10^8))/(3.99*10^-19)=4.98*10^-7m=498nm