134.Atomic Calculation From the Bohr Model

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Number 134

Atomic Calculations from the Bohr Model Bohr knew that the hydrogen atom consisted of a negatively charged electron orbiting a positively charged proton but… although atoms are small there was a very BIG problem.

The electron was behaving like a wave once it was part of an atom. Even better, to satisfy Bohr’s condition that the electron didn’t radiate energy away as it ‘orbited’, the electron wave had to be a standing or stationary wave.

The electron was orbiting in a circle. That meant it was accelerating. But accelerating charges always give out electromagnetic radiation. So the electron should lose energy.

Exam Hint: These are the shapes of the first few standing waves. It is a good idea to learn them.

Bohr took the dramatic step of saying that the electron in an atom could only be in certain orbits and couldn’t be anywhere else. In addition, when the electrons were in these orbits they didn’t radiate any energy at all. Furthermore, he said he could calculate the radius of the closest possible orbit. It was r = 5.3 × 10-11 m, which became known as the Bohr Radius. It might seem obvious - the electron is a small particle. But in the case of the atom the electron didn’t behave like a particle. For a start when an electron fell towards a proton and ended up at the Bohr radius it did not turn its potential energy into kinetic energy, it gave out a photon of energy equal to the loss in potential energy (13.6eV)

A Prince to the Rescue! Prince Louis de Broglie, a dozen years later, found himself thinking that waves and particles were really the same thing. de Broglie (pronounced ‘de Brooey’) came up with an equation which linked particles and waves. It is: λ = h/p λ is the wavelength of the ‘wave’ in metres (m) h is Planck’s constant, (6.63 × 10-34 Js) p is the momentum of the ‘particle’ (kgms-1)

Bohr had calculated other radii where the electron could be orbiting and still not give out any energy. These radii turned out to be where the next possible standing waves could wrap themselves around the proton.

Worked Example 1 What is the ‘wavelength’ of the electron in the Bohr orbit?

Exam Hint: Be careful when you say exactly which standing wave shapes occupy the Bohr orbits.

Momentum = mass × velocity. So λ = h/mv As h, m, and v were known:

λ =

Only the waves with whole numbers of wavelengths are stable. So only the second and fourth shapes shown above correspond to actual orbits.

6.63 × 10-34 = 3.32 × 10-10 m (9.1 × 10-31 × 2.19 × 106 )

Now consider this: Take the Bohr radius. Multiply it by 2π. This gives the circumference of the Bohr orbit. Put the numbers in:

It was only standing waves that could exist. If the waves weren’t standing waves they would not be stable and so would give out their energy and cease to exist.

2πr = 2 × 3.14 × 5.3 × 10-11 = 3.32 × 10-10 m But this was exactly the same as the ‘wavelength’ of the electron that was in that orbit! The conclusion was inescapable.

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Physics Factsheet

134. Atomic Calculations from the Bohr Model

The result was that the emission spectrum of hydrogen consisted of lines which showed a definite pattern. This pattern could be explained by using the idea of electrons falling back to the ground state from various higher energy levels (or of standing waves returning to their simplest allowed form).

The standing waves corresponded to the energy levels found by experiment as shown below.

- 0.85 eV E4 E3

- 1.51 eV

- 0.54 eV - 0.85 eV - 1.51 eV

E2

3rd set of lines (I-R)

- 3.39 eV E1

- 3.39 eV

2nd set of lines (visible)

- 13.6 eV

- 13.6 eV

E0 1st set of lines (u-v)

Now it was clear. The electron wasn’t a particle at all once it was orbiting around the proton. It was acting as a standing wave.

Taken directly from the above diagram, the emission spectrum would look like this.

Worked Example 2 Remember that an electron falling towards a proton from a large distance away will finally settle into an orbit whose radius is the Bohr radius. Or perhaps we should say it will become a standing wave in its simplest allowed pattern. Either way it still has to get rid of 13.6eV of potential energy. (see diagram above). It does this by radiating away a photon of exactly this energy.

Increasing wavelength

E = hc/ λ = 13.6eV = 13.6 × 1.6 × 10-19 = 2.18 × 10-18 J

u-v

hc (6.63 × 10-34 × 3 × 108) so λ = -18 = 2.18 × 10 2.18 × 10-18 -8 Answer = 9.1 × 10 m

visible

I-R

Practice Questions 1. The Bohr model of a hydrogen atom assumes that an electron e is in a circular orbit around a proton P.

This is in the ultraviolet part of the electromagnetic spectrum. By experiment on hydrogen atoms, light of exactly this wavelength was observed to be given out when a far-away electron fell towards a proton to make a complete hydrogen atom. The electron, or wave, is not stable at higher energy levels. So it quickly, in a fraction of a second, returns to its lowest possible energy level. The energy difference is radiated away from the atom in the form of a photon. Exam Hint: Be careful which way the electron moves. If it drops down energy levels then a photon is emitted. But an electron can also absorb a photon of exactly the right energy and move up energy levels. Make sure you make clear to the examiner that you know whether the photon is being absorbed or emitted.

In the ground state the orbit has a radius of 5.3 × 10-11 m. At this separation the electron is attracted to the proton by a force of 8.1 × 10-8 N. (a) State what is meant by the ground state.

Worked example 3

(b) (i) Show that the speed of the electron in this orbit is about 2.2 × 106ms-1 (mass of an electron = 9.1 × 10-31 kg)

If the electron fell from the -3.39eV level (otherwise known as the first excited state) back to the -13.6eV level (the ground state) then the energy difference between these levels is 13.6 – 3.39 = 10.21eV. This is:

(ii) Calculate the de Broglie wavelength of an electron travelling at this speed. (Planck constant = 6.63 × 10-34 Js)

10.21 × 1.6 × 10-19 = 1.63 × 10-18 J and this is the exact energy of the photon emitted.

(iii) How many waves of this wavelength fit the circumference of the electron orbit?

Using E = hc/λ we can calculate the wavelength of this photon. It is: λ = hc/E = (6.63 × 10-34 × 3 × 108) / 1.63 × 10-18 = 1.22 × 10-7 m.

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Physics Factsheet

134. Atomic Calculations from the Bohr Model

4. The diagram shows some of the energy levels of a Thallium atom.

2. The energy level diagram shown below is a very simplified diagram for the hydrogen atom.

0

First excited state

0 eV –3.4 eV Energy/eV

Ground state

Ionisation

–1.6

–5.5

–13.6 eV

A moving electron with a kinetic energy of 12.5 eV collides with an atom of hydrogen and causes the electron in the ground state to be raised to the first excited state.

–10.4

(a) Calculate the kinetic energy of the moving electron after the collision (in eV)

(a) Calculate the ionisation energy in Joules for an electron in the –10.4 eV ground state.

(b) Calculate the wavelength of the photon emitted when the electron in the first excited state returns to the ground state.

(b) A neutron of kinetic energy 9.4 eV collides with a Thallium atom. As a result, an electron moves from the –10.4 eV level to the –1.6 eV level. What is the kinetic energy in eV of the neutron after the collision?

3. A muon is a particle which was discovered in a particle accelerator. It is a ‘heavy electron’. It has exactly the same properties as an electron but its mass is 208 times greater.

(c) A transition between which two energy levels in the Thallium atom will give rise to an emission line of wavelength close to 320 nm? Draw this transition on the diagram above.

Scientists have made an ‘atom’ in which the muon orbits a proton. The muon energy levels for this ‘atom’ has been discovered.

Answers

0 eV

1. (b) (i) 2.17 × 106 ms-1

–312 eV

(ii) 3.3 × 10-10 m (iii) Exactly One!

–703 eV

2. (a) 2.3 eV (b) 1.22 × 10-7 m 3. (a) 2810 eV (b) 4.42 × 10-10 m (d) 1.17 × 10-13 m

–2810 eV

4. (a) 1.664 × 10-18 J (b) 0.6 eV (c) photon energy = 3.88 eV

Ground state

(a) Write down the ionisation energy of this atom. (b) Hence calculate the maximum possible wavelength of a photon which, when absorbed, would be able to ionise this ‘atom’. (c) To which part of the electromagnetic spectrum does this wavelength belong? (d) Calculate the de Broglie wavelength of a muon travelling at 10% of the speed of light. Acknowledgements: This Physics Factsheet was researched and written by Paul Dutton The Curriculum Press,Bank House, 105 King Street,Wellington, Shropshire, TF1 1NU Physics Factsheets may be copied free of charge by teaching staff or students, provided that their school is a registered subscriber. No part of these Factsheets may be reproduced, stored in a retrieval system, or transmitted, in any other form or by any other means, without the prior permission of the publisher. ISSN 1351-5136

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