Answer:
B. 2.18×10^−18J
Explanation:
Transition from n = 1 to n =[infinity]
Ionization Energy = ?
Energy of a photon is directly proportional to its frequency as described by the Planck - Einstein relation:
E = h⋅ν
Here
E is the energy of the photon
h is Planck's constant, equal to 6.626⋅10−34J s
calculating the wavelength of the emission line that corresponds to an electron that undergoes a n=1 → n=∞ transition in a hydrogen atom.
This transition is part of the Lyman series (hydrogen spectral series of transitions and resulting ultraviolet emission lines of the hydrogen atom as an electron goes from n ≥ 2 to n = 1 (where n is the principal quantum number), the lowest energy level of the electron.) and takes place in the ultraviolet part of the electromagnetic spectrum.
The wavelength λ of the emission line in the hydrogen spectrum is given by Rydberg equation for the hydrogen atom;
1 / λ = R ⋅ ( 1 /n²₁ − 1 / n²₂)
Where:
λ is the wavelength of the emitted photon (in a vacuum)
R is the Rydberg constant, equal to 1.097⋅10^7 m−1
n₁ represents the principal quantum number of the orbital that is lower in energy
n₂ represents the principal quantum number of the orbital that is higher in energy
In this problem;
n₁ = 1
n₂ = ∞
At higher and higher values the expression tends to zero until at n=∞. as the value of n₂ increases, the value of 1 / n²₂ decreases. When n=∞, you can say that;
1 / n²₂ → 0
Upon solving;
1 / λ = R ⋅ (1 /n²₁ - 0)
1 / λ= R ⋅ 1 /n²₁
Since n₁ = 1 this becomes:
1 / λ = R
1 / λ = 1.097 × 10^7
λ= 9.116 × 10^−8 m
We now use the following formular to find the frequency and hence the corresponding energy:
c = ν * λ
ν = c / λ = 3×10^8 / 9.116×10^−8 = 3.291 × 10^15 s−1
Now we can use the Planck expression:
E = h * ν
E = 6.626×10^−34 × 3.291×10^15 = 2.18×10^−18J
