Respuesta :
Answer:
a) λ = 435 nm , c) c) λ = 4052 nm, d) λ= 95 nm
Explanation:
A) To carry out this excitation, the energy of the laser must be greater than or equal to the energy of the transition of the hydrogen atom, whose states of energy are described by the Bohr model.
En = -13,606 / n² [eV]
therefore the energy of the transition is
ΔE = E₅ -E₂
ΔE = 13.606 (1 / n₂² - 1 / n₅²)
ΔE = 13.606 (1/2² - 1/5²)
ΔE = 2,85726 eV
now let's use Planck's equation
E = h f
the speed of light is related to wavelength and frequencies
c = λ f
f = c /λ
E = h c /λ
λ = h c / E
let's reduce the energy to the SI system
E = 2,85726 eV (1.6 10⁻¹⁹ J / 1 eV) = 4.5716 10⁻¹⁹ J
let's calculate
λ = 6,626 10⁻³⁴ 3 10⁸ / 4,5716 10⁻¹⁹
λ = 4.348 10⁺⁷ m (10⁹ nm / 1 m)
λ = 435 nm
B) photon emission processes from this state with n = 5 to the base state n = 1, can give transition
initial state n = 5
final state n = 4
ΔE = 13.606 (1/4² - 1/5²)
ΔE = 0.306 eV
λ = h c / E
λ = 4052 nm
n = 5
final ΔE (eV) λ (nm)
level
4 0.306 4052
3 0.9675 1281
2 2,857 435
1 13.06 95
n = 4
3 0.661 1876
2 2,551 486
1 11,905 104
n = 3
2 1.89 656
1 12.09 102.5
n = 2
1 10.20 121.6
c) λ = 4052 nm
d) λ= 95 nm
