Question

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

Answer 1

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