Grade 12

Advanced

8 min

Lesson 7 of 12

Atomic Models and Photon Emission

Not Started

Connecting quantum mechanics to the structure of the atom and the production of spectral lines.

Why does a neon sign glow bright red, while a mercury lamp shines a ghostly blue? The answer lies in the fact that atoms are not messy clouds, but precisely tuned 'quantum instruments' that can only play specific notes of light.

Learning Objectives

1.By the end of this lesson, you will be able to calculate the energy and wavelength of photons emitted during electron transitions.
2.You'll understand how the Bohr model explains the quantization of electron orbits.
3.You'll be able to relate discrete atomic energy levels to the unique emission and absorption spectra of elements.

The Bohr Model and Quantization

In 1913, Niels Bohr proposed that electrons do not spiral into the nucleus but exist in stationary states or orbits. Unlike a satellite that can orbit at any distance, an electron is quantized; it can only occupy specific, allowed energy levels. Bohr postulated that the angular momentum of an electron is an integer multiple of

\frac{h}{2\pi}

. This means an electron's energy is fixed while in an orbit, and it only changes energy by 'jumping' between these levels. For a Hydrogen atom, the energy of a level

n

is given by:

E_n = -\frac{13.6 \text{ eV}}{n^2}

where

n

is the principal quantum number (

n=1, 2, 3...

Quick Check

If an electron is in the ground state (n=1), can it emit a photon to drop to a lower energy level?

Consider the definition of the 'ground state'.

Answer

No, the ground state is the lowest possible energy level (n=1); there are no lower states to transition into.

Calculate the energy of an electron in the $n=3$ level (the second excited state) of a Hydrogen atom.

1. Identify the formula: $E_n = -\frac{13.6}{n^2} \text{ eV}$ . 2. Substitute $n=3$ : $E_3 = -\frac{13.6}{3^2}$ . 3. Calculate: $E_3 = -\frac{13.6}{9} \approx -1.51 \text{ eV}$ .

Photon Emission and Absorption

When an electron transitions from a high energy level (

E_{high}

) to a lower one (

E_{low}

), it must lose energy. This energy is released as a single photon. The energy of this photon is exactly equal to the difference between the two levels:

\Delta E = E_{high} - E_{low}

Using the Planck-Einstein relation, we can find the frequency (

f

) or wavelength (

\lambda

) of the light:

\Delta E = hf = \frac{hc}{\lambda}

where

h

is Planck's constant (

6.63 \times 10^{-34} \text{ J}\cdot\text{s}

) and

c

is the speed of light (

3 \times 10^8 \text{ m/s}

). Note:

1 \text{ eV} = 1.6 \times 10^{-19} \text{ J}

An electron in a Hydrogen atom drops from $n=3$ to $n=2$ . Find the wavelength of the emitted photon.

1. Find $\Delta E$ in eV: $\Delta E = E_3 - E_2 = (-1.51 \text{ eV}) - (-3.40 \text{ eV}) = 1.89 \text{ eV}$ . 2. Convert to Joules: $1.89 \times 1.6 \times 10^{-19} \text{ J} = 3.024 \times 10^{-19} \text{ J}$ . 3. Use $\lambda = \frac{hc}{\Delta E}$ : $\lambda = \frac{(6.63 \times 10^{-34})(3 \times 10^8)}{3.024 \times 10^{-19}}$ . 4. Result: $\lambda \approx 6.58 \times 10^{-7} \text{ m}$ (or $658 \text{ nm}$ , which is red light).

Quick Check

Does a transition from n=4 to n=1 result in a higher or lower frequency photon than a transition from n=2 to n=1?

Look at how the energy gap changes as the difference between n-values increases.

Answer

Higher frequency, because the energy gap (ΔE) is larger, and energy is directly proportional to frequency (E=hf).

Atomic Fingerprints: Spectral Lines

Because the energy levels are discrete, an atom can only emit or absorb specific wavelengths. This creates a line spectrum. An emission spectrum consists of bright lines on a dark background, produced when excited electrons drop to lower states. Conversely, an absorption spectrum shows dark lines on a continuous rainbow, produced when electrons absorb specific photons to jump to higher states. These lines act as a 'fingerprint' for elements, allowing astronomers to determine the composition of distant stars just by analyzing the light they emit.

The Lyman series involves transitions ending at $n=1$ . Calculate the shortest possible wavelength (the series limit) for Hydrogen.

1. The shortest wavelength corresponds to the highest energy transition. 2. This occurs when an electron falls from $n = \infty$ (ionization) to $n=1$ . 3. $\Delta E = E_{\infty} - E_1 = 0 - (-13.6 \text{ eV}) = 13.6 \text{ eV}$ . 4. Convert to Joules: $13.6 \times 1.6 \times 10^{-19} = 2.176 \times 10^{-18} \text{ J}$ . 5. $\lambda = \frac{hc}{\Delta E} = \frac{1.989 \times 10^{-25}}{2.176 \times 10^{-18}} \approx 9.14 \times 10^{-8} \text{ m}$ (Ultraviolet).

Key Takeaways

1Electrons exist in quantized orbits; they cannot exist between these defined energy levels.
2Photon energy is the difference between levels: $\Delta E = E_{initial} - E_{final}$ .
3Spectral lines are unique to each element because every element has a unique set of energy levels.
4The energy of a Hydrogen level is $E_n = -\frac{13.6}{n^2} \text{ eV}$ .

Quick Check

Multiple Choice

What happens to the energy of an electron as the principal quantum number $n$ increases?

Multiple Choice

Which transition in a Hydrogen atom would emit a photon with the longest wavelength?

True/False

An absorption spectrum is produced when electrons move from a higher energy level to a lower energy level.

What's Next?

Review Tomorrow

In 24 hours, try to sketch the first four energy levels of Hydrogen and calculate the energy of the n=4 level without looking at your notes.

Practice Activity

Research the 'Balmer Series' and identify which visible colors correspond to the transitions ending at n=2.

Browse

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Atomic Models and Photon Emission

Learning Objectives

The Bohr Model and Quantization

Photon Emission and Absorption

Atomic Fingerprints: Spectral Lines

Key Takeaways

Quick Check

What's Next?

Atomic Models and Photon Emission

Learning Objectives

The Bohr Model and Quantization

Photon Emission and Absorption

Atomic Fingerprints: Spectral Lines

Key Takeaways

Quick Check

What's Next?