Question

Light with a wavelength of 614.5 nm looks orange. What is the energy, in joules, per photon of this orange light? What is the energy in eV (1 eV = 1.602 × 10^−19 J)?

a) 3.22 × 10^−19 J; 2.01 eV
b) 2.01 × 10^−19 J; 3.22 eV
c) 4.05 × 10^−19 J; 2.52 eV
d) 2.52 × 10^−19 J; 4.05 eV

Answer 1

The energy, in joules, per photon of orange light with a wavelength of 614.5 nm is 3.22 × 10-19 J, and the energy in eV is 2.01 eV.

Step-by-step explanation:

To calculate the energy, in joules, per photon of orange light with a wavelength of 614.5 nm, we can use the equation E = hc/λ, where E is the energy, h is Planck's constant (6.63 × 10-34 J·s), c is the speed of light (3.00 × 108 m/s), and λ is the wavelength in meters. The energy, in joules, per photon of orange light with a wavelength of 614.5 nm is 3.22 × 10-19 J, and the energy in eV is 2.01 eV.

Converting the wavelength to meters, we have 614.5 nm = 6.145 × 10-7 m. Plugging these values into the equation, we get E = (6.63 × 10-34 J·s)(3.00 × 108 m/s)/(6.145 × 10-7 m) = 3.22 × 10-19 J per photon.

To convert this energy to electron volts (eV), we can use the conversion factor 1 eV = 1.602 × 10-19 J. Therefore, the energy in eV is (3.22 × 10-19 J)/(1.602 × 10-19 J/eV) = 2.01 eV per photon. Therefore, the correct answer is option a) 3.22 × 10-19 J; 2.01 eV.