Answer:
4.74*10^-7 m
Explanation:
TO find the wavelength of the photon you calculate the energy of the photon emitted in a harmonic oscillator:
[tex]E_{m-n}=\hbar \omega(m+\frac{1}{2})-\hbar \omega(n+\frac{1}{2})=\hbar \omega (m-n)\\\\\omega=\sqrt{\frac{k}{m}}\\\\E_{m-n}=\hbar \sqrt{\frac{k}{m}}(m-n)[/tex]
m: initial state = 3
n: final state = 1
k = 3.6N/m
By replacing the values of m for the electron, m,n and ħ you obtain:
[tex]E_{m-n}=(1.055*10^{-34}Js)\sqrt\frac{3.6N/m}{9.11*10^{-31}kg}}(3-1)=4.19*10^{-19}J[/tex]
Furthermore, this energy is equivalent to the expression:
[tex]E_{3-1}=\hbar \omega=\hbar 2\pi \nu=2\pi \hbar\frac{c}{\lambda}\\\\\lambda=\frac{2\pi \hbar c}{E_{3-1}}[/tex]
By replacing you obtain:
[tex]\lambda=\frac{2\pi (1.055*10^{-34}Js)(3*10^8m/s)}{4.19*10^{-19}J}=4.74*10^{-7}m[/tex]
hence, the wavelength of the photon is 4.74*10^-7 m
The wavelength of the emission if the net force on the electron behaves as though it has a spring constant of 3.6 N/m - [tex]4.74\times10^-7 m[/tex]
The energy of the photon emitted in a harmonic oscillator:
[tex]E_{m-n}=\hbar \omega(m+\frac{1}{2})-\hbar \omega(n+\frac{1}{2})=\hbar \omega (m-n)\\\\\omega=\sqrt{\frac{k}{m}}\\\\E_{m-n}=\hbar \sqrt{\frac{k}{m}}(m-n)[/tex]
m: initial state = 3
n: final state = 1
k = 3.6N/m
By replacing the values of m for the electron, m, n and ħ you obtain:
[tex]E_{3-1}=\hbar \omega=\hbar 2\pi \nu=2\pi \hbar\frac{c}{\lambda}\\\\\lambda=\frac{2\pi \hbar c}{E_{3-1}}[/tex]
Thus, the wavelength of the photon is [tex]4.74\times10^-7 m[/tex]
Learn more:
https://brainly.com/question/17026166
