Question

The rays of the Sun that cause tanning and burning are in the ultraviolet portion of the electromagnetic spectrum. These rays are categorized by wavelength: So-called UV-A radiation has wavelengths in the range of 320-380 nm, whereas UV-B radiation has wavelengths in the range of 290-320 nm.

(a) Calculate the energy of a mole of 360 nm photons. (answer in Kj/mol)

An AM radio station broadcasts at 1140 kHz, and its FM partner broadcasts at 101.0 MHz. Calculate and compare the energy of photons emitted by these radio stations.

(b) E_FM / E_AM

Answer 1

a)

Energy of a photon is given by following expression:

E = hc/λ

Where, h = plank’s constant (6.63 x 10^-34 m²kg/s)

c = velocity of light (3 x 10⁸ m/s)

λ = wavelength (360 nm)

E = (6.63 x 10^-34 m²kg/s) x (3 x 10⁸ m/s) / (360 nm)

Convert nm into m by multiplying 10^-9

E = (6.63 x 10^-34 m²kg/s) x (3 x 10⁸ m/s) / (360 x 10^-9 m)

E = 0.05525 x 10^-17 m²kg/s²

E = 0.05525 x 10^-20 kJ

Therefore, energy of 1 photon is 0.05525 x 10^-20 kJ. One mole of photons contain 6.026 x 10²³ photons. Therefore calculate energy of one mole of photons as follows:

Energy of 1 mole of photons = 6.026 x 10²³ mol^-1 x 0.05525 x 10^-20 kJ

Energy of 1 mole of photons = 0.3329 x 10³ kJ mol^-1

Energy of 1 mole of photons = 332.9 kJ mol^-1

Therefore, energy of a mole of 360 nm photons is 332.9 kJ mol^-1.

--------------------------------------------------------------------------------------------------------------

b)

Expression of energy of photon in terms of frequency is given as follows:

E = hv

Where, v = frequency (Hz)

Calculate energy of AM photon having frequency 1140 kHz

E = (6.63 x 10^-34 m²kg/s) (1140 kHz)

Convert kHz into Hz

E = (6.63 x 10^-34 m²kg/s) (1140 x 10³ Hz)

E = 7558.2 x 10^-31 J

Therefore, energy of a photon from AM wave is 7558.2 x 10^-31 J

Similarly, calculate energy of photon of FM station having frequency 101.0 MHz.

E = (6.63 x 10^-34 m²kg/s) (101.0 MHz)

E = (6.63 x 10^-34 m²kg/s) (101.0 x 10⁶ Hz)

E = 669.63 x 10^-28 J

Therefore, energy of a photon from FM wave is 669.63 x 10^-28 J

Ratio of energy of energy of FM photon (EFM) to energy of AM photon (EAM) is calculated as follows:

EFM/EAM = (669.63 x 10^-28 J) / (7558.2 x 10^-31 J)

EFM/EAM = 88.6

Therefore, Ratio of energy of energy of FM photon (EFM) to energy of AM photon (EAM) is 88.6