Answer:
the wavelength, in nm, of the photon is 487.5 nm
Explanation:
Given:
n = 4 (excited)
n = 2 (relaxes)
Question: Calculate the wavelength, in nm, λ = ?
First, it is important to calculate the energy of the electron when it excited and then when it relaxes.
[tex]E_{1} =\frac{-13.6}{n^{2} } =\frac{-13.6}{4^{2} } =-0.85eV[/tex] (excited)
[tex]E_{2} =\frac{-13.6}{2^{2} } =-3.4eV[/tex] (relaxes)
The change of energy
ΔE = E₁ - E₂=-0.85 - (-3.4) = 2.55 eV = 4.08x10⁻¹⁹J
For a photon, the wavelength emitted
[tex]\lambda =\frac{hc}{delta(E)}[/tex]
Here
h = Planck's constant = 6.63x10⁻³⁴J s
c = speed of light = 3x10⁸m/s
Substituting values:
[tex]\lambda =\frac{6.63x10^{-34}*3x10^{8} }{4.08x10^{-19} } =4.875x10^{-7} m=487.5nm[/tex]
