Answer:
ΔE = 7.559 eV
, λ = 1,645 10⁻⁷ m
Explanation:
For this exercise we can use the Bohr model for ionized atom with only one free electron,
r_n = n² a₀ / Z
E_n = -13,606 Z² / n²
Where a₀ is the Bohr radius of the hydrogen atom (a₀ = 0.0529 nm), Z is the atomic number of the atom under study and 13.606 eV is the energy of the ground state of Hydrogen.
In our case the Helium atom has two protons Z = 2
let's calculate the quantum number and the energy of each orbit
r_n = 0.30 nm
n₁ = √ (r_n Z / a₀)
n₁ = √ (0.30 2 / 0.0529)
Note that we do not have to reduce the radius since they are all in nanometers
n₁ = 3.3
since n is an integer we approximate it to
n₁ = 3
r_n = 0.20 nm
n₂ = √ (0.2 2 / 0.0529)
n₂ = 2.7
To approximate this value we must assume that there could be some error in the medicinal radio,
n₂ = 2
having the quantum numbers of the two radius we can calculate their energy
E₃ = - 13,606 2²/3²
E₃ = - 6.047 eV
E₂ = -13.606 2²/2²
E₂ = -13.606 eV
the energy of the emitted photon is
ΔE = E₃ - E₂
ΔE = -6.047 + 13.606
ΔE = 7.559 eV
You do not indicate in the exercise if you want the energy or the wavelength of the photon,
to find the wavelength We use the Planck relation
E = h f
c = λ f
E = h c /λ
λ = h c / E
we must reduce the energy to the SI system
E = 7.559 ev (1.6 10⁻¹⁹ J / 1eV) = 12.09 10⁻¹⁹ J
λ = 6.63 10⁻³⁴ 3 10⁸ / 12.09 10⁻¹⁹
λ = 1,645 10⁻⁷ m
