Answer :
A classical frequency of in a harmonic oscillator is(f)= 6.15×10¹⁴ Hz
The wavelength of the photon ([tex]\lambda[/tex])= 488 nm.
How to calculate the values of the frequency and the wavelength?
To calculate the values of the frequency and the wavelength we are using the formula is,
f= [tex]\frac{E}{h}[/tex]
[tex]\lambda = \frac{c}{f}[/tex]
Here we are given,
E= energy of photon.
f= classical frequency of in a harmonic oscillator.
[tex]\lambda[/tex] = The wavelength of the photon.
To calculate the energy we are using a formula,
E = 13.6× [tex](\frac{1}{n_{2}^{2} } -\frac{1}{n_{1}^{2} })[/tex]
n₂= the second excited state = 4
n₁= the ground state = 2
Now we put the values in the equation we get,
E= 13.6× [tex](\frac{1}{4^{2} } -\frac{1}{2^{2} })[/tex]
Or, E= 2.55 eV
h= Planck constant.
= 6.627 ×10³⁴ J.S
c= Speed of light.
=3×10⁸ m/s
Now we put the known values to deduce the frequency,
f= [tex]\frac{E}{h}[/tex]
Or, f= [tex]\frac{2.55\times 1.6\times 10^{-19}}{6.627\times 10{-34}}[/tex]
Or, f= 6.15×10¹⁴ Hz
Now from the above calculation we can say that, The frequency of the a harmonic oscillator is(f)= 6.15×10¹⁴ Hz
Now we put the values in the above equation and deduce the wavelength,
[tex]\lambda = \frac{c}{f}[/tex]
Or, [tex]\lambda = \frac{3\times 10^8}{6.15\times 10^{14}}[/tex]
Or,[tex]\lambda[/tex]= 4.88×10⁻⁷ m= 488nm.
From the above calculation we can conclude that, The wavelength of the photon is ([tex]\lambda[/tex])= 488 nm.
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