Answer:1870nm
Explanation:
The energy of the hydrogen atom in the quantum state n is E = - (13.6 eV) / n^2. The ground state, when n = 1 and E = - 13.6 eV.
after absorption,the energy of atom is -13.6 + 12.75 = -0.85 eV.
quantum number for this energy state is
-0.85 = -13.6/n^2
n^2 = 13.6/0.85 = 16
n =4.
The next-lower energy orbit has energy
E = -13.6/3^2 = -1.511
Then transition emits a photon of energy -0.85-(-1.511) = 0.661 eV.
you would use
Landa = (planck's constant)(speed of light) / energy
λ = hc/E
or in your case, λ = (6.63 x 10-34 * 3 x 10^8) / (0.661 / 6.24 x 10^18)
to get λ = 1870 nm
The wavelength of the photon emitted in this quantum jump is equal to 1870 nm.
First of all, we would determine the energy of this atom after absorption as follows:
Energy = -13.6 + 12.75
Energy = -0.85 eV.
Next, we would determine the quantum number for the above energy state after absorption:
-0.85 = -13.6/n²
n² = 13.6/0.85
n = √16
n = 4.
Also, the energy for the next lower-energy orbit is given by:
E = -13.6/3²
E = -1.511 Joules.
For this transition, it emits a photon of energy:
E = -0.85-(-1.511)
E = 0.661 eV.
Now, we can calculate the wavelength by Planck-Einstein equation:
λ = hc/E
λ = (6.63 × 10⁻³⁴ × 3 × 10₈)/(0.661/6.24 × 10¹⁸)
Wavelength, λ = 1870 nm.
Read more on photon energy here: brainly.com/question/9655595
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