Respuesta :
Answer:
Explanation:
94.91 nm = 1237.5 / 94.91 eV
= 13.038 eV
Energy of ground state = - 13.6 eV
Energy of higher state to which it jumped
= - 13.6 + 13.038
= - .562 eV
If n be the orbit to which it jumped
13.6 / n² = .562
n = 5 ( approx )
b )
1281 nm = 1237.5 / 1281
= .966 eV
Energy of second photon = 13.038 - .966
= 12.072 eV
Wavelength of second photon
= 1237.5 / 12.072
= 12.51 nm
b ) Energy of intermediate level
= - 0.562 - .966 eV
= - 1.528 eV
If it be n th level
13.6 / n² = 1.528
n = 3
A) The higher level that the electron reached from ground state is; n = 5
B) The intermediate level that the electron reached from ground state is; n = 3
C) The wavelength of the second photon emitted is; λ = 103 nm
A) We are given;
Wavelength of photon absorbed by ground state H atoms; λ_g = 94.91 nm = 94.91 × 10⁻⁹ m
Formula to get the higher level is Rydberg's formula;
1/λ = R(1/n₁² - 1/n₂²)
where;
R is rydberg constant = 1.097 × 10⁷ m⁻¹
Thus;
1/(94.91 × 10⁻⁹) = 1.097 × 10⁷(1/1² - 1/n₂²)
0.9605 = 1 - 1/n₂²
1/n₂² = 1 - 0.9605
1/n₂² = 0.0395
n₂ = √(1/0.0395)
n₂ ≈ 5
B) We want to find the intermediate level where wavelength = 1281 nm = 1281 × 10⁻⁹ m
Thus;
1/(1281 × 10⁻⁹) = 1.097 × 10⁷(1/n₂² - 1/5²)
0.0712 = 1/n₂² - ¹/₂₅
1/n₂² = 0.0712 + ¹/₂₅
1/n₂² = 0.1112
n₂ = √(1/0.1112)
n₂ ≈ 3
C) Formula for energy of photon is;
E = hc/λ
where;
h is Planck's constant = 6.626 × 10⁻³⁴ m².kg/s
c is speed of light = 3 × 10⁸ m/s
Thus;
E = (6.626 × 10⁻³⁴ × 3 × 10⁸)/(1281 × 10⁻⁹)
E = 1.552 × 10⁻¹⁹ J
The energy at ground state is;
E = (6.626 × 10⁻³⁴ × 3 × 10⁸)/(94.91 × 10⁻⁹)
E = 20.944 × 10⁻¹⁹ J
Thus;
Energy of second photon = (20.944 × 10⁻¹⁹) - (1.552 × 10⁻¹⁹)
Energy of second photon = 19.352 × 10⁻¹⁹ J
Wavelength of second photon emitted is;
λ = hc/E
λ = (6.626 × 10⁻³⁴ × 3 × 10⁸)/19.352 × 10⁻¹⁹
λ = 103 nm
Read more about energy of a photon at; https://brainly.com/question/7464909
