Determine the energy of 2.00 mol of photons for each kind of light.

Question

Determine the energy of 2.00 mol of photons for each kind of light.

Part A infrared radiation  1540 (nm)

Part B visible light 520 (nm)

Part C ultraviolet radiation  140 (nm)

Answer 1

Since you are given the wavelength of each type of light, you can use E = hC/λ

E = energy =?

h = Planck's constant = 6.626x10^-34 J*sec

C = speed of light = 3x10⁸ m/sec

λ = wavelength in m = 1540 nm x 1 m/10⁹nm = 1.54x10^-6 m for part A

E = (6.626x10^-34 Jsec)(3x10⁸ m/sec)/1.54x10-6 m = 12.9x10^-20 J = 1.3x10^-19 J per photon.

1.3x10^-19 J/photon x 6.02x10²³ photons/mole x 2 moles = 1.6x10⁵ J = energy of 2 moles of photons in part A

You should be able to do the other wavelengths the same way by substituting the appropriate nm into the equation. (HINT: as the wavelength decreases, the energy E will increase).