Question

A 14N14N molecule is vibrating in its ground state with a spring constant of 2233 N m−1. What is the wavelength of the photon which excited this system to the first excited state? Use scientific notation to express your answer (X.XXE-X).

Answer 1

Answer:

(6.675 × 10⁶) m

Explanation:

A body vibrating in simple harmonic motion has its frequency related to the spring constant and mass of the body through the relation

f = (1/2π) √(k/m)

k = 2233 N/m

m = mass of one molecule of Nitrogen = 28.0 g = 0.028 kg

f = (1/2π) √(2233/0.028)

f = 44.945 Hz

And the photons that'll excite the molecule to the next state must at least have this frequency (f = 44.945 Hz)

For waves, the velocity (v), frequency (f) and wavelength (λ) are related through the relation

v = fλ

v = speed of light (since it's a photon) = (3 × 10⁸) m/s

λ = (v/f) = (3 × 10⁸)/44.945 = 6674824.8 m = (6.675 × 10⁶) m = (6.68 E6) m