Syllabus Edition

First teaching 2014

Last exams 2024

|

Quantization of Angular Momentum (DP IB Physics: HL)

Revision Note

Katie M

Author

Katie M

Last updated

Quantization of Angular Momentum

  • Angular momentum is a property of any spinning or rotating body, very similar to linear momentum
    • In linear motion, momentum is the product of mass and velocity
  • In rotational motion the momentum is the product of moment of inertia and angular speed

  • Angular momentum is a vector, this means:
    • The magnitude is equal to the momentum of the particle times its radial distance from the centre of its circular orbit
    • The direction of the angular momentum vector is normal to the plane of its orbit with the direction being given by the corkscrew rule

12-1-7-quantization-ib-hl

Angular momentum acts at right angles to the direction of rotation

  • Niels Bohr proposed that the angular momentum, L, of an electron in an energy level is quantised in integer multiples of fraction numerator h over denominator 2 straight pi end fraction
  • Where:
    • n = an integer (n = 1, 2, 3...)
    • h = Planck’s constant
  • Sometimes h may be written as the reduced Planck's constant, ħ, which is equal to
ħ equals fraction numerator h over denominator 2 straight pi end fraction

  • Hence the angular momentum for an electron in a circular orbit is constant
  • De Broglie proposed that an electron with momentum p = mv has a wavelength λ given by

lambda equals h over p

  • For an electron moving in a straight line, the matter wave takes a familiar wave shape consisting of peaks and troughs
    • Although the electron itself isn't oscillating up and down, only the matter wave

12-1-7-de-broglie-matter-wave-for-an-electron-1-ib-hl

de Broglie matter wave for an electron moving in a straight line at constant speed

  • For the same electron moving in a circle, the matter wave still has a sinusoidal shape but is wrapped into a circle

12-1-7-de-broglie-matter-wave-for-an-electron-2-ib-hl

de Broglie matter wave for an electron moving in a circular orbit at constant speed

  • As the electron continues to orbit in a circle two possibilities may occur:

 1. On completing one oscillation, the waves overlap in phase

    • The waves will continue in phase over many orbits giving rise to constructive interference and a standing wave 

12-1-7-de-broglie-matter-wave-for-an-electron-4-ib-hl

de Broglie matter wave where 3λ is less than the orbits circumference

2. On completing one oscillation, the waves overlap but they are not in phase

    • In other words, peak overlaps with peak, trough with trough
    • This means that where the waves overlap, destructive interference occurs and as a result, no such electron orbit is allowed

12-1-7-de-broglie-matter-wave-for-an-electron-3-ib-hl

de Broglie matter wave where n = 3. Here the circumference of the circular orbit is 3λ

  • Hence, the circumference of the orbit open parentheses 2 pi r close parentheses must equal an integer number of wavelengths open parentheses n lambda close parentheses for a standing wave to form:

n lambda equals 2 pi r

  • Using the de Broglie relation, lambda equals h over straight p:

n h over straight p equals 2 pi r

  • Since momentum is equal to p equals m v:

n fraction numerator h over denominator m v end fraction equals 2 pi r

  • Rearranging for angular momentum, m v r:

n fraction numerator straight h over denominator 2 straight pi end fraction equals m v r

Bohr Condition

  • The Bohr Condition is given by the relation:

n fraction numerator straight h over denominator 2 straight pi end fraction space equals space m v r

  • Where:
    • n = energy level 
    • h = Planck's constant (J s)
    • m = mass (kg)
    • = velocity (m s–1)
    • r = radius (m)
  • The condition essentially states that an electron can only move in fixed orbits
  • Neils Bohr and Ernest Rutherford found that atomic orbits are only allowed when:
    • The angular momentum of the electron is an integer multiple of fraction numerator h over denominator 2 straight pi end fraction
  • The Bohr condition also leads to the restriction of the electron radius to certain values 
  • Another implication is that the discrete or quantised energy of the electron follows this rule:

begin mathsize 14px style E equals negative fraction numerator 13.6 over denominator n squared end fraction end styleeV

Worked example

Determine the velocity of the electron in the first Bohr orbit of the hydrogen atom (n = 1).

You may use the following values:

  • Mass of an electron = 9.1 × 10−31 kg
  • Radius of the orbit = 0.529 × 10−10 m
  • Planck's constant = 6.63 × 10−34 kg m2 s-1

Step 1: List the known quantities

    • Mass of an electron, m = 9.1 × 10−31 kg
    • Radius of the orbit, r = 0.529 × 10−10 m
    • Planck's constant, h = 6.63 × 10−34 kg m2 s-1

Step 2: Write the Bohr Condition equation and rearrange for velocity, v

size 14px v size 14px space size 14px equals size 14px space fraction numerator size 14px n size 14px h over denominator size 14px 2 size 14px pi size 14px space size 14px space size 14px cross times size 14px space size 14px m size 14px r end fraction

Step 3: Substitute in the values and calculate, v

size 14px v size 14px space size 14px equals size 14px space fraction numerator size 14px 1 size 14px space size 14px cross times size 14px space size 14px left parenthesis size 14px 6 size 14px. size 14px 63 size 14px cross times size 14px space size 14px 10 to the power of size 14px minus size 14px 34 end exponent size 14px right parenthesis over denominator size 14px 2 size 14px pi size 14px space size 14px space size 14px cross times size 14px space size 14px left parenthesis size 14px 9 size 14px. size 14px 1 size 14px space size 14px cross times size 14px space size 14px 10 to the power of size 14px minus size 14px 31 end exponent size 14px right parenthesis size 14px space size 14px cross times size 14px space size 14px left parenthesis size 14px 0 size 14px. size 14px 529 size 14px space size 14px cross times size 14px space size 14px 10 to the power of size 14px minus size 14px 10 end exponent size 14px right parenthesis end fraction size 14px space size 14px equals size 14px space size 14px 2 size 14px. size 14px 2 size 14px space size 14px cross times size 14px space size 14px 10 to the power of size 14px 6 size 14px space size 14px m size 14px space size 14px s to the power of size 14px minus size 14px 1 end exponent

Step 4: Write the final answer

The velocity of the electron = 2.2 × 106 m s-1

You've read 0 of your 5 free revision notes this week

Sign up now. It’s free!

Join the 100,000+ Students that ❤️ Save My Exams

the (exam) results speak for themselves:

Did this page help you?

Katie M

Author: Katie M

Expertise: Physics

Katie has always been passionate about the sciences, and completed a degree in Astrophysics at Sheffield University. She decided that she wanted to inspire other young people, so moved to Bristol to complete a PGCE in Secondary Science. She particularly loves creating fun and absorbing materials to help students achieve their exam potential.