Question

Confirm the statement in the that the range of photon energies for visible light is 1.63 to 3.26 eV, given that the range of visible wavelengths is 380 to 760 nm.

1. True
2. False

Answer 1

Using the provided conversion that hc equals 1240 eV nm and the equation E = (hc)/λ, the calculated range of photon energies for visible light from 380 to 760 nm indeed translates to 1.63 to 3.26 eV, confirming the statement as True.

Step-by-step explanation:

To confirm the statement that the range of photon energies for visible light is 1.63 to 3.26 eV, we use the fact that the energy (E) of a photon can be calculated from its wavelength (λ) using the equation E = (hc)/λ, where h is Planck's constant (6.626 x 10-34 Joule seconds) and c is the speed of light (3 x 108 meters per second). To work with electron volts (eV) and nanometers (nm), we use the conversion that hc = 1240 eV nm, as provided in the text.

For the shortest wavelength in the visible range (380 nm):

E = 1240 eV nm / 380 nm = 3.26 eV

For the longest wavelength in the visible range (760 nm):

E = 1240 eV nm / 760 nm = 1.63 eV

This calculation confirms that the range of photon energies for visible light is indeed 1.63 to 3.26 eV, which means the statement is True.