## Physics A: Problem Set 24: Atomic Spectra

### recommended reading

Barron's Let's Review: | 13.5 Models of the Atom |

physics.info: | n/a |

Wikipedia: | Bohr model, Energy level |

Khan Academy: | Bohr's model of the hydrogen atom |

YouTube: | Why is glass transparent? |

### classwork

- In a mercury atom, as an electron moves from energy level i to energy level a, a single photon is emitted.
- Determine the emitted photon's energy.
- What kind of electromagnetic radiation is this?

You need a reference table that labels the energy levels of mercury with letters. I only know of one. The question didn't state a preferred unit so I am going to provide answers in both electronvolts and joules.

*E*= |*E*−*E*_{0}|*E*= 10.38 eV − 1.56 eV*E*= 8.82 eVConvert electronvolts to joules by multiplying by the elementary charge.

(8.82 eV)(1.60 × 10

^{−19}C) = 1.41 × 10^{−18}JConsult a reference source to determine the type of radiation. Depending on the source, you'll need either the frequency or wavelength of the emitted photon. Use Planck's equation in some form to determine one of these quantities. Use the units of your choice. Just be consistent. Here's four possible ways to do the computation — first using SI units and then using acceptable non SI units. No matter how you slice it, this photon is in the ultraviolet part of the electromagnetic spectrum.

First using SI units.

*f*=*E**h**f*=1.41 × 10 ^{−18}J6.63 × 10 ^{−34}J·s*f*= 2.13 × 10^{15}Hzλ = *hc**E*λ = (6.63 × 10 ^{−34}J·s)(3.00 × 10^{8}m/s)1.41 × 10 ^{−18}Jλ = 1.41 × 10 ^{−7}mNow using acceptable non SI units.

*f*=*E**h**f*=8.82 eV 4.14 × 10 ^{−15}eVs*f*= 2.13 × 10^{15}Hzλ = *hc**E*λ = 1240 eVnm 8.82 eV λ = 141 nm

### homework

- A photon with a frequency of 5.02 × 10
^{14}Hz is absorbed by an excited hydrogen atom. This causes the electron to be ejected from the atom, forming an ion.- Calculate the energy of the photon in joules.
- Determine the energy of the photon in electron volts.
- What is the number of the lowest energy level (closest to the ground state) of a hydrogen atom that contains an electron that would be ejected by the absorption of this photon?
- What is the kinetic energy of the ejected electron in electron volts?
- What is the kinetic energy of the ejected electron in joules?
- How fast is the electron moving?

Use Planck's equation to compute the energy of a photon from its frequency.

*E*=*hf**E*= (6.63 × 10^{−34}J·s)(5.02 × 10^{14}Hz)*E*= 3.33 × 10^{−19}JConvert joules to electronvolts by dividing by the elementary charge.

3.33 × 10 ^{−19}J= 2.08 eV 1.60 × 10 ^{−19}CYou need to consult a reference table of some sort for this. An electron in the third energy level has an energy of −1.51 eV. Our photon has more than enough energy to eject that electron. The energy of the second energy level is −3.40 eV. Our photon will not be able to eject that electron.

Add the energy of the photon to the energy of the electron — and by add I mean subtract, since the energy of the electron is negative. The ejected electron will carry along this excess energy as kinetic energy.

2.08 eV − 1.51 eV = 0.57 eV

- Convert electronvolts to joules by multiplying by the elementary charge.
(0.57 eV)(1.60 × 10

^{−19}C) = 9.12 × 10^{−20}J Rearrange the kinetic energy equation to compute the speed of the electron.

*v*=√ 2 *K**m**v*=√ 2(9.12 × 10 ^{−20}J)9.11 × 10 ^{−31}kg*v*= 447,000 m/sSpeedy.