Quantization of angular momentum IB DP Physics Study Notes - 2025 Syllabus
Quantization of angular momentum IB DP Physics Study Notes
Quantization of angular momentum IB DP Physics Study Notes at IITian Academy focus on specific topic and type of questions asked in actual exam. Study Notes focus on IB Physics syllabus with Students should understand
that emission and absorption spectra provide evidence for discrete atomic energy levels
that photons are emitted and absorbed during atomic transitions
that the frequency of the photon released during an atomic transition depends on the difference in energy level as given by $E = hƒ$
that emission and absorption spectra provide information on the chemical composition
the discrete energy levels in the Bohr model for hydrogen as given by $E = − \frac{13 . 6}{ n^2} eV$
that the existence of quantized energy and orbits arise from the quantization of angular momentum in the Bohr model for hydrogen as given by $mvr =\frac{ nh }{2π}$ .
Standard level and higher level: 6 hours
Additional higher level: 3 hours
- IB DP Physics 2025 SL- IB Style Practice Questions with Answer-Topic Wise-Paper 1
- IB DP Physics 2025 HL- IB Style Practice Questions with Answer-Topic Wise-Paper 1
- IB DP Physics 2025 SL- IB Style Practice Questions with Answer-Topic Wise-Paper 2
- IB DP Physics 2025 HL- IB Style Practice Questions with Answer-Topic Wise-Paper 2
Atomic spectra and atomic energy states – review
When a low-pressure gas in a tube is subjected to a voltage the gas ionizes and emits light.
We can analyze that light by looking at it through a spectroscope.
∙A spectroscope acts similar to a prism in that it separates the incident light into its constituent wavelengths.
∙For example, heated barium gas will produce an emission spectrum that looks like this:
FYI
∙ Heating a gas and observing its light is how we produce and observe atomic spectra
∙The fact that the emission spectrum is discontinuous tells us that atomic energy states are quantized.
- Now we know that light energy is carried by a particle called a photon.
- If a photon of just the right energy strikes a hydrogen atom, it is absorbed by the atom and stored by virtue of the electron jumping to a new energy level:
- The electron jumped from the n = 1 state to the n = 3 state.
- We say the atom is excited.
- When the atom de-excites the electron jumps back down to a lower energy level.
- When it does, it emits a photon of just the right energy to account for the atom’s energy loss during the electron’s orbital drop.
- The electron jumped from the n = 3 state to the n = 2 state.
- We say the atom is de-excited, but not quite in its lowest, or ground state.
PRACTICE:
A spectroscopic examination of glowing hydrogen shows the presence of a 434 nm blue emission line.
(a) What is its frequency?
(b) What is the energy (in J and eV) of each of its blue-light photons?
▶️Answer/Explanation
SOLUTION(a)
Use λf = c with c = 3.00 x 10⁸ m s⁻¹ and λ = 434 x 10⁻⁹ m:
(434 x 10⁻⁹)f = 3.00 x 10⁸
f = (3.00 x 10⁸) / (434 x 10⁻⁹)
= 6.91 x 10¹⁴ Hz
SOLUTION(b)
Use \(E = hf\):
\[E = (6.63 \times 10^{-34})(6.91 \times 10^{14})\]
\[= 4.58 \times 10^{-19} \text{ J}\]
\[= (4.58 \times 10^{-19} \text{ J}) \left(\frac{1 \text{ eV}}{1.60 \times 10^{-19} \text{ J}}\right)\]
\[= 2.86 \text{ eV}.\]
PRACTICE:
Calculate the radius of a hydrogen atom in its ground state ($n = 1$).
▶️Answer/Explanation
SOLUTION:
$r_{1}=\frac{1^{2}h^{2}}{4\pi^{2}ke^{2}m}$
$=\frac{(6.63\times10^{-34})^{2}}{4\pi^{2}(8.99\times10^{9})(1.60\times10^{-19})^{2}(9.11\times10^{-31})}$
$=5.31\times10^{-11} \text{ m.}$
PRACTICE:
Given $E = -\frac{ke^2}{2r_n}$, show that
$E_n = -\frac{13.6}{n^2}$ ($E$ in eV) hydrogen energy levels
▶️Answer/Explanation
SOLUTION:
$E_n = -\frac{ke^2}{2r_n} = -(ke^2)(\frac{4\pi^2ke^2m}{2n^2h^2}) = -\frac{2\pi^2k^2e^4m}{n^2h^2}$
$= \frac{2\pi^2(8.99\times10^9)^2(1.60\times10^{-19})^4(9.11\times10^{-31})}{(6.63\times10^{-34})^2n^2}$
$= -\frac{(2.1668\times10^{-18} \text{J})(\frac{1 \text{ eV}}{1.60\times10^{-19} \text{ J}})}{n^2}$
$= -\frac{13.6}{n^2} \text{ eV.}$