To solve, use the formula of energy level of harmonic oscillator.
E_n = (n + 1/2)hf
where
E_n is the energy level of harmonic oscillator in Joules
n is the quantum level
h is the Planck's constant (6.623 xx 10^(-34) Js)
and f is the frequency of oscillator.
To be able to apply this formula, convert the given energy to Joules. Take note that 1eV = 1.602xx10^(-19)J .
E_n=5.45 eV * (1.602xx10^(-19)J)/(1eV)
E_n = 8.7309xx10^(-19) J
Plug-in this value of En to the formula of energy level of harmonic oscillator.
E_n = (n+1/2)hf
8.8309 xx10^(-19) J= (3+1/2)(6.623 xx 10^(-34) Js)f
8.7309 xx10^(-19) J=(2.31805 xx10^(-33) Js) f
Then, isolate the f.
f = (8.7309 xx 10^(-19)J)/(2.31805xx10^(-33)Js)
f=3.76648476xx10^14/ sec
f=3.76648476 xx 10^14Hz
So the frequency of the oscillator is 3.76648476xx10^14Hz .
To determine the angular frequency, apply the formula:
omega = 2pif
omega =2pi * (3.76648476xx10^14 Hz)
omega=2.366552170 xx10^15 rad/s
Rounding off to two decimal places, it becomes:
omega =2.37 xx10^15 rad/s
Therefore, the angular frequency of harmonic oscillator is 2.37xx10^15 radian per second.
Sunday, March 1, 2015
The energy of a particle in the n = 3 excited state of a harmonic oscillator. Potential is 5.45 eV. What is the classical angular frequency of oscillation of this particle?
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