E=hf=6.63•10⁻³⁴•1.7•10¹³=1.13•10⁻²⁰ J=
= (1.13•10⁻²⁰/1.6•10⁻¹⁹) eV = 0.07 eV
E=(n+0.5)hf
n=(E/hf) - 0.5 =
=(1.6•10⁻¹⁹/6.63•10⁻³⁴•1.7•10¹³) – 0.5 ≃14
A LiBr molecule oscillates with a frequency of 1.7×1013 Hz.
(a) What is the difference in energy in eV between allowed oscillator states?
(b) What is the approximate value of n for a state having an energy of 1.0 eV?
1 answer
1 answer