Question

A harmonic oscillator with spring constant, k, and mass, m, loses 3 quanta of energy, leading to the emission of a photon.

a. What is the energyon in terms of k adm
b. If the oscillator is a bonded atom with k = 15 N/m and m = 4 × 10-26 kg, what is the frequency (Hz) of the emitted photon? (Note: the energy of a photon is Ephoton= hf)
c. In which region of the electromagnetic spectrum (x-ray, visible, microwave, etc.) does this photon belong?

Answer 1

Answer: (a). E = 3.1656×10³⁴ √k/m

(b). f = 9.246 × 10¹² Hz

(c). Infrared region.

Step-by-step explanation:

From Quantum Theory,

The energy of a proton is proportional to the frequency, from the equation;

E = hf

where E = energy in joules

h = planck's constant i.e. 6.626*10³⁴ Js

f = frequency

(a). from E = hf = 1 quanta

f = ω/2π

where ω = √k/m

consider 3 quanta of energy is lost;

E = 3hf = 3h/2π × √k/m

E = (3×6.626×10³⁴ / 2π) × √k/m

E = 3.1656×10³⁴ √k/m

(b). given from the question that K = 15 N/m

and mass M = 4 × 10⁻²⁶ kg

To get the frequency of the emitted photon,

Ephoton =hf = 3h/2π × √k/m (h cancels out)

f = 3h/2π × √k/m

f = 3h/2π × (√15 / 4 × 10⁻²⁶ )

f = 9.246 × 10¹² Hz

(c). The region of electromagnetic spectrum, the photon belongs to is the Infrared Spectrum because the frequency ranges from about 3 GHz to 400 THz in the electromagnetic spectrum.