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The Bohr Model of Hydrogen (HL) (HL IB Physics)

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Katie M

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Katie M

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The Bohr Model of Hydrogen

  • Hydrogen is the simplest atom in existence, making it ideal for experiments investigating the nature of electron energy levels
  • Line spectra produced by hydrogen atoms showed that
    • Electrons are able to jump, or transition, between specific energy levels producing specific energy photons
    • Different transitions can be categorised into series, or families, of lines
  • The Lyman series converges on the ground state (n = 1) for electrons
    • The Balmer series converges on the second energy level (n = 2)
    • The Ritz-Paschen converges on the third energy level (n = 3) and so on

The transitions observed in hydrogen line spectra can be classed into series

  • The Lyman series photons will have the highest energies since they have the shortest wavelength
    • These transitions tend to produce ultraviolet photons
  • The Pfund series photons will have the lowest energies since they have the longest wavelength
    • These transitions tend to produce infrared photons

7-1-3-electron-jumps-in-the-hydrogen-spectrum

Representing the electron jumps in the hydrogen spectra

  • The finding of these electron transitions helped scientists to understand how electrons work to produce photons of specific wavelength and energy
  • This led to the development of the Bohr model of hydrogen, which states that
    • Electrons can only move in fixed orbits
    • The orbital radius of electrons is restricted to certain values
  • The discrete energy of the transitions in the Bohr model for hydrogen are described by the equation:

E space equals space minus fraction numerator 13.6 over denominator n squared end fraction space eV

  • Where
    • E = photon energy (J)
    • n = an integer 1, 2, 3 etc. to describe the energy level of an atom

Worked example

Determine the frequency of an emitted photon from a hydrogen atom when an electron makes a transition between levels n = 4 and n = 2.

Answer:

Step 1: List the known quantities

  • Transition between n = 4 and n = 2
  • Planck’s constant, h = 6.63 × 10–34 J s
  • Electronvolt, eV = 1.6 × 10–19 J

Step 2: Determine an equation for the change in energy ΔE

E subscript n space equals space minus fraction numerator 13.6 space e V over denominator n squared end fraction

increment E space equals space E subscript 4 space minus space E subscript 2

increment E space equals space open parentheses negative fraction numerator 13.6 space e V over denominator 4 to the power of 4 end fraction close parentheses space minus space open parentheses negative fraction numerator 13.6 space e V over denominator 2 to the power of 4 end fraction close parentheses 

Step 3: Calculate the change in energy, in eV, for the photon using the given equation

increment E space equals space minus 13.6 open parentheses 1 over 4 squared space minus space 1 over 2 squared close parentheses space equals space 2.55 space e V

Step 4: Rearrange the photon energy equation for frequency f

E space equals space h f space space space space space rightwards double arrow space space space space space f space equals space E over h

Step 5: Substitute the known values into the equation for frequency

f space equals space fraction numerator 2.55 cross times open parentheses 1.6 space cross times space 10 to the power of negative 19 end exponent close parentheses over denominator 6.63 space cross times space 10 to the power of negative 34 end exponent end fraction space equals space 6.15 space cross times space 10 to the power of 14 space Hz

Quantisation 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, which 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 Planck's constant over 2π:

L space equals space n fraction numerator h over denominator 2 straight pi end fraction

  • Where:
    • n = an integer (n = 1, 2, 3...)
    • h = Planck’s constant
  • 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 space equals space 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 is

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 n = 3. Here the circumference of the circular orbit is 3λ

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 3λ is less than the orbit's circumference

  • Hence, the circumference of the orbit open parentheses 2 straight 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 space equals space 2 straight pi r

  • Using the de Broglie relation:

lambda space equals h over p

n open parentheses h over p close parentheses space equals space 2 straight pi r

  • Since momentum is equal to space p space equals space m v:

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

  • Rearranging for angular momentum open parentheses L space equals space m v r close parentheses:

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

  • This is known as the Bohr Condition 
  • Where:
    • n = integer number of energy level
    • h = Planck's constant (J s)
    • m = mass of an electron (kg)
    • = velocity of electron (m s–1)
    • r = radius of orbit (m)

Worked example

Determine the velocity of the electron in the first Bohr orbit of the hydrogen atom.

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

Answer:

Step 1: List the known quantities

  • First orbital level, n = 1
  • 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

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

Step 3: Substitute the values in and calculate the velocity v

v space equals space fraction numerator 1 space cross times space open parentheses 6.63 cross times 10 to the power of negative 34 end exponent close parentheses over denominator 2 straight pi cross times open parentheses 9.1 cross times 10 to the power of negative 31 end exponent close parentheses open parentheses 0.529 cross times 10 to the power of negative 10 end exponent close parentheses end fraction

Step 4: Write the final answer

Velocity of an electron (n = 1):  v = 2.2 × 106 m s−1

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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.