1. Bohr, Emissions, and Spectra
DR. MIOY T. HUYNH
YALE UNIVERSITY
CHEMISTRY 161
FALL 2018

2. ELECTRONS BEHAVE VERY MUCH LIKE LIGHT!
Q: What is an electron?
Is it a wave that carries energy?
Is it a negatively charged particle?
A: It behaves as both a wave and a particle.
ELECTRONS BEHAVE VERY MUCH LIKE LIGHT!

3. The Bohr Model: Quantization
Energy is discrete!
Energy is not continuous!
excited electron
“Electrons should move around the nucleus but only in prescribed orbits. When jumping from one orbit to another with lower energy, a light quantum is emitted. Energy is transferred only in certain well defined quantities.”
Bohr's theory could explain why atoms emitted light in fixed wavelengths and colors.

4. The Bohr Model: Quantization
If we imagine the nucleus at the center, then:
n = 1 has the lowest energy.
n = 2 has the second lowest energy.
The energies get higher the farther we get from the nucleus.
The spacing between states also gets smaller!
n = 4
n = 3
ENERGY
n = 2
n = 1
NUCLEUS

5. Electron Transitions ENERGY
Now, say we have an electron in the n = 2 state with energy E2.
This electron can drop to the lower energy state n = 1, which has E1.
In doing so, this transition also emits a photon of light with a wavelength equal to:
n = 5
n = 4
n = 3
ENERGY
n = 2
1 λ n 2 → n 1 = × 10 −2 n m − 𝑛 1 2 − 1 𝑛 2 2
n = 1
NUCLEUS

6. Electron Transitions ENERGY
Now, say we have an electron in the n = 3 state with energy E3.
This electron can drop to the lower energy state n = 1, which has E1.
In doing so, this transition also emits a photon of light with a wavelength equal to:
n = 5
n = 4
n = 3
ENERGY
n = 2
1 λ n 3 → n 1 = × 10 −2 n m − 𝑛 1 2 − 1 𝑛 3 2
n = 1
NUCLEUS

7. Electron Transitions ENERGY
Now, say we have an electron in the n = 3 state with energy E3.
This electron can drop to the lower energy state n = 1, which has E1.
In doing so, this transition also emits a photon of light with a wavelength equal to:
This electron can also drop to the lower energy state n = 2 and emit a photon of light with a wavelength equal to:
n = 5
n = 4
n = 3
ENERGY
n = 2
1 λ n 3 → n 1 = × 10 −2 n m − 𝑛 1 2 − 1 𝑛 3 2
n = 1
NUCLEUS
1 λ n 3 → n 2 = × 10 −2 n m − 𝑛 2 2 − 1 𝑛 3 2

8. Electron Transitions ENERGY Generally:
So, if we have an electron in the state n = 3 with energy E3.
This electron can drop to either of the two lower energy states n = 1 or n = 2.
These transitions would emit photons of light with wavelengths equal to:
n = 5
n = 4
n = 3
ENERGY
n = 2
1 λ n 3 → n 1 = × 10 −2 n m − 𝑛 1 2 − 1 𝑛 3 2
n = 1
1 λ n 3 → n 2 = × 10 −2 n m − 𝑛 2 2 − 1 𝑛 3 2
NUCLEUS
Generally:
1 λ = × 10 −2 n m − 𝑛 initial 2 − 1 𝑛 final and ΔE= −2.178 × 10 −18 J 1 𝑛 final 2 − 1 𝑛 initial 2

9. Hydrogen atoms are excited into the n = 4 state.
In what region of the electromagnetic spectrum would you expect for the radiation with the lowest energy?

10. Hydrogen atoms are excited into the n = 4 state.
In what region of the electromagnetic spectrum would you expect for the radiation with the lowest energy?
An electron in the n = 4 state could transition to the n = 3, n = 2, and n =1 states.
n = 5
n = 4
n = 3
ENERGY
n = 2
n = 1
NUCLEUS

11. Hydrogen atoms are excited into the n = 4 state.
In what region of the electromagnetic spectrum would you expect for the radiation with the lowest energy?
An electron in the n = 4 state could transition to the n = 3, n = 2, and n =1 states.
n = 5
n = 4
n = 3
ENERGY
n = 2
n = 1
NUCLEUS

12. Hydrogen atoms are excited into the n = 4 state.
In what region of the electromagnetic spectrum would you expect for the radiation with the lowest energy?
An electron in the n = 4 state could transition to the n = 3, n = 2, and n =1 states.
These transitions would emit photons of light with wavelengths equal to:
n = 5
n = 4
n = 3
ENERGY
n = 2
n = 1
NUCLEUS

13. Hydrogen atoms are excited into the n = 4 state.
In what region of the electromagnetic spectrum would you expect for the radiation with the lowest energy?
An electron in the n = 4 state could transition to the n = 3, n = 2, and n =1 states.
These transitions would emit photons of light with wavelengths equal to:
n = 5
n = 4
n = 3
1 λ 4→1 = × 10 −2 n m − −
ENERGY
n = 2
1 λ 4→2 = × 10 −2 n m − −
1 λ 4→3 = × 10 −2 n m − −
n = 1
NUCLEUS

14. Hydrogen atoms are excited into the n = 4 state.
In what region of the electromagnetic spectrum would you expect for the radiation with the lowest energy?
An electron in the n = 4 state could transition to the n = 3, n = 2, and n =1 states.
These transitions would emit photons of light with wavelengths equal to:
n = 5
n = 4
n = 3
1 λ 4→1 = × 10 −2 n m − − =1.028 × 10 −2 n m −1
ENERGY
n = 2
1 λ 4→2 = × 10 −2 n m − − =2.057 × 10 −3 n m −1
1 λ 4→3 = × 10 −2 n m − − =5.333 × 10 −4 n m −1
n = 1
NUCLEUS

15. Hydrogen atoms are excited into the n = 4 state.
In what region of the electromagnetic spectrum would you expect for the radiation with the lowest energy?
An electron in the n = 4 state could transition to the n = 3, n = 2, and n =1 states.
These transitions would emit photons of light with wavelengths equal to:
n = 5
n = 4
n = 3
1 λ 4→1 = × 10 −2 n m − − =1.028 × 10 −2 n m −1 ⇒ λ 4→1 =97.23 nm
ENERGY
n = 2
1 λ 4→2 = × 10 −2 n m − − =2.057 × 10 −3 n m −1 ⇒ λ 4→2 =486.2 nm
1 λ 4→3 = × 10 −2 n m − − =5.333 × 10 −4 n m −1 ⇒ λ 4→3 =1875 nm
n = 1
NUCLEUS

16. Hydrogen atoms are excited into the n = 4 state.
In what region of the electromagnetic spectrum would you expect for the radiation with the lowest energy?
An electron in the n = 4 state could transition to the n = 3, n = 2, and n =1 states.
These transitions would emit photons of light with wavelengths equal to:
Recall that longer λ’s correspond to lower energy waves (radiation), so the electron transition from the n = 4 to n = 3 states would yield the lowest energy radiation, which emits in the the
n = 5
n = 4
n = 3
1 λ 4→1 = × 10 −2 n m − − =1.028 × 10 −2 n m −1 ⇒ λ 4→1 =97.23 nm
ENERGY
n = 2
1 λ 4→2 = × 10 −2 n m − − =2.057 × 10 −3 n m −1 ⇒ λ 4→2 =486.2 nm
1 λ 4→3 = × 10 −2 n m − − =5.333 × 10 −4 n m −1 ⇒ λ 4→3 =1875 nm
n = 1
NUCLEUS

17. Hydrogen atoms are excited into the n = 4 state.
In what region of the electromagnetic spectrum would you expect for the radiation with the lowest energy?
An electron in the n = 4 state could transition to the n = 3, n = 2, and n =1 states.
These transitions would emit photons of light with wavelengths equal to:
Recall that longer λ’s correspond to lower energy waves (radiation), so the electron transition from the n = 4 to n = 3 states would yield the lowest energy radiation, which emits in the the INFRARED REGION.
n = 5
n = 4
n = 3
1 λ 4→1 = × 10 −2 n m − − =1.028 × 10 −2 n m −1 ⇒ λ 4→1 =97.23 nm
ENERGY
n = 2
1 λ 4→2 = × 10 −2 n m − − =2.057 × 10 −3 n m −1 ⇒ λ 4→2 =486.2 nm
1 λ 4→3 = × 10 −2 n m − − =5.333 × 10 −4 n m −1 ⇒ λ 4→3 =1875 nm
n = 1
NUCLEUS

18. Hydrogen atoms are excited into the n = 4 state.
In what region of the electromagnetic spectrum would you expect for the radiation with the lowest energy?
An electron in the n = 4 state could transition to the n = 3, n = 2, and n =1 states.
These transitions would emit photons of light with wavelengths equal to:
Recall that longer λ’s correspond to lower energy waves (radiation), so the electron transition from the n = 4 to n = 3 states would yield the lowest energy radiation, which emits in the the INFRARED REGION.
Q: Can you calculate the energy of this emission?
n = 5
n = 4
n = 3
1 λ 4→1 = × 10 −2 n m − − =1.028 × 10 −2 n m −1 ⇒ λ 4→1 =97.23 nm
ENERGY
n = 2
1 λ 4→2 = × 10 −2 n m − − =2.057 × 10 −3 n m −1 ⇒ λ 4→2 =486.2 nm
1 λ 4→3 = × 10 −2 n m − − =5.333 × 10 −4 n m −1 ⇒ λ 4→3 =1875 nm
n = 1
NUCLEUS

19. Hydrogen atoms are excited into the n = 4 state.
In what region of the electromagnetic spectrum would you expect for the radiation with the lowest energy?
An electron in the n = 4 state could transition to the n = 3, n = 2, and n =1 states.
These transitions would emit photons of light with wavelengths equal to:
Recall that longer λ’s correspond to lower energy waves (radiation), so the electron transition from the n = 4 to n = 3 states would yield the lowest energy radiation, which emits in the the INFRARED REGION.
Q: Can you calculate the energy of this emission? A: ΔE4→3 = –1.059 × 10–19 J
n = 5
n = 4
n = 3
1 λ 4→1 = × 10 −2 n m − − =1.028 × 10 −2 n m −1 ⇒ λ 4→1 =97.23 nm
ENERGY
n = 2
1 λ 4→2 = × 10 −2 n m − − =2.057 × 10 −3 n m −1 ⇒ λ 4→2 =486.2 nm
1 λ 4→3 = × 10 −2 n m − − =5.333 × 10 −4 n m −1 ⇒ λ 4→3 =1875 nm
n = 1
NUCLEUS