It is possible to determine the ionization energy for hydrogen using the Bohr equation. Calculate the ionization energy for an atom of hydrogen, making the assumption that ionization is the transition from n = 1 to n =[infinity]a. -2.18 x 10^-18 Jb. +2 .18 x 10^-18 Jc. +4.59 x 10-^18 Jd. -4.59 x 10-^18 Je. +4.36 x 10^-18 J

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

120,630,890

step-by-step explanation:

110+10=120,620+10=630,880+10=890