Answer:
λ₁ = 87.5 10⁻¹² m
, λ₂ = 2.175 10⁻¹⁰ m, E₂ = 5.8 10³ eV
Explanation:
In this case you can use the law of conservation of energy, all the energy of the electron is converted into energized emitted photons
Let's reduce to the SI system
E₀ = 20 10³ eV (1.6 10⁻¹⁹ J / 1eV) = 3.2 10⁻¹⁵ J
Δλ = 1.30 A = 0.13 nm = 0.13 10⁻⁹ m
Ef = E₁ + E₂
E₀ = Ef
E₀ = E₁ + E₂
The energy can be found with the Planck equation
E = h f
c = λ f
f = c / λ
E = hc / λ
They indicate that the wavelength of the second photon is
λ₂ = λ₁ +0.130 10⁻⁹
We replace
E₀ = hv / λ₁ + hc / ( λ₁ + 0.130 10⁺⁹)
E₀ / hv = 1 / λ₁ + 1 / ( λ₁ + 0.13 10⁻⁹)
3.2 10⁻¹⁵ / (6.63 10⁻³⁴ 3 10⁸) = ( λ₁ + 0.13 10⁻⁹ + λ₁) / λ₁ ( λ₁ + 0.13 10⁻⁹)
1.6 10¹⁰ ( λ₁² +0.13 10⁻⁹ λ₁) = 2 λ₁ + 0.13 10⁻⁹
λ₁² + 0.13 10⁻⁹ λ₁ = 1.25 10⁻¹⁰ λ₁ + 8.125 10⁻²¹
λ₁² + 0.005 10⁻⁹ λ₁ = 8.125 10⁻²¹
λ₁² + 5 10⁻¹² λ₁ - 8.125 10⁻²¹ = 0
Let's solve the second degree equation
λ₁ = [-5 10⁻¹² ±√((5 10⁻¹²)² + 4 8.125 10⁻²¹)] / 2
λ₁ = [-5 10⁻¹² ±√(25 10⁻²⁴ +32.5 10⁻²¹)] / 2 = [-5 10⁻¹² ±√ (32525 10⁻²⁴)] / 2
λ₁ = [-5 10⁻¹² ± 180 10⁻¹²] / 2
λ₁ = 87.5 10⁻¹² m
λ₂ = -92.5 10⁻¹² m
We take the positive wavelength
The wavelength of the photons is
λ₁ = 87.5 10⁻¹² m
λ₂ = λ₁ + 0.13 10⁻⁹
λ₂ = 87.5 10⁻¹² + 0.13 10⁻⁹
λ₂ = 0.2175 10⁻⁹ m = 2.175 10⁻¹⁰ m
The energy after the first deceleration is
E₂ = E₀ –E₁
E₂ = E₀ –hc / λ₁
E₂ = 3.2 10⁻¹⁵ - 6.63 10⁺³⁴ 3 10⁸ / 87.5 10⁻¹²
E₂ = 3.2 10⁻¹⁵ - 2.27 10⁻¹⁵
E₂ = 0.93 10⁻¹⁵ J
E₂ = 0.93 10⁻¹⁵ J (1 eV / 1.6 10⁻¹⁹ J)
E₂ = 5.8 10³ eV
