Different energy levels

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1 Different energy levels In the microscopic world energy is discrete

2 Review Atomic electrons can only exist in certain discrete energy levels Light is made of photons When e s lose energy they give out light à emission spectrum When light is absorbed it gives energy to the e s à absorpcon spectrum E = nhf à energy is quanczed, n=1, 2, 3 corresponding to the energy levels At higher frequency (shorter wavelength) photon has more energy Frequency of emined or absorbed light is fixed by E b/w energy levels

3 Topic 7: Atomic, nuclear and particle physics 7.1 Discrete energy and radioactivity Transitions between energy levels The first 7 energy levels for hydrogen are shown here: The energy levels are labeled from the lowest to the highest as n = 1 to n = 7 in the picture. n is called the principal quantum number and goes all the way up to infinity ( )! In its ground state or unexcited state, hydrogen s single electron is in the 1 st energy level (n = 1):

4 Hydrogen energy levels Values are given in ev All values are negative PE of two electrons are zero only when they are at infinite separation Electron has moved from infinity into its orbit Proton electron has lost some energy and since it was zero to start with, it is now negative Level 1 is most negative à has smallest amount of energy

5 Relationship between energy levels and frequency of photons E upper E lower =hf n=3 E 3 = 1.51eV n=2 E 2 = 3.40eV Electron is dropping from n=3 to n=2 E = (- 3.40) = +1.89eV n=3 n=2

6 Hydrogen Spectrum Increasing λ The generalization of the allowed energy for hydrogen is given by E n = 13.6/ n 2 ev N is the quantum number an can be any positive integer

7 Example Calculate the energy of the ground state and first exited state of the hydrogen atom SoluCon E 1 = 13.6/ 1 2 ev= 13.6eV E 2 = 13.6/ 2 2 ev= 3.4eV

8 Example Ultra violet part of the EM spectrum Calculate the frequency and wavelength of the photon emined when an electron makes a quantum jump from n = 3 to the ground state in the hydrogen atom Solu%on Frequency is in Hertz = 1/s, need to convert ev into J E 1 = 13.6/ 1 2 ev= 13.6eV x1.6 x = 2.18x J E 3 = 13.6/ 3 2 ev= 1.51eVx1.6 x = 2.42x J E= E 3 E 1 =hf f= 2.42x ( 2.18x )/6.6x =2.9x Hz λ= c/f = 3.0x 10 8 /2.9x =1.0x 10 7 m

9 Energy levels for Hydrogen A simple mathemaccal model was established in 1885 by Johann Balmer to predict the wavelength of the visible spectrum of hydrogen λ=b( n 2 / n ) λ= wavelength b = constant = x 10 5 cm n = quantum number

10 Topic 7: Atomic, nuclear and particle physics 7.1 Discrete energy and radioactivity Transitions between energy levels The previous energy level diagram was NOT to scale. This one is. Note that none of the energy drops of the other series overlap those of the Balmer series, and thus we cannot see any of them. But we can still sense them! Second Excited State First Excited State Ground State n = n = 5 n = 4 n = 3 n = 2 n = 1 Balmer Series (Visible) Lyman Series (UV) [ SUNBURN ] FYI Transition energy is measured in ev because of the tiny amounts involved ev ev ev ev Paschen Series (IR) [ HEAT ] ev ev

11 Topic 7: Atomic, nuclear and particle physics 7.1 Discrete energy and radioactivity Transitions between energy levels EXAMPLE: An electron jumps from energy level n = 3 to energy level n = 2 in the hydrogen atom. (a) What series is this de-excitation in? SOLUTION: Find it on the diagram: This jump is contained in the Balmer Series, and produces a visible photon.

12 Rydberg series 1/λ =R( 1/ m 2 1/ n 2 ) R = Rydberg constant = 1.096x 10 7 m 1 m=3 m=2 m=1

13 Limitations of the model It failed to predict the spectra for atoms that have two or more electrons It could not predict the relacve brightness of the lines It could not account for two or more closely spaced lines such as the pair found in the yellow region of the visible spectrum of sodium It was unable to explain the bonding of atoms in molecules or in solids or in liquids Bohr could not find any evidence for his postulates, which blended classical and quantum Physics without any juscficacon

14 Topic 7: Atomic, nuclear and particle physics 7.1 Discrete energy and radioactivity Solving problems involving atomic spectra PRACTICE: A spectroscopic examination of glowing hydrogen shows the presence of a 434 nm blue emission line. (c) What are the energy levels associated with this photon? SOLUTION: Because it is visible use the Balmer Series with E = ev. Note that E 2 E 5 = = ev. Thus the electron jumped from n = 5 to n = 2.

15 Topic 7: Atomic, nuclear and particle physics 7.1 Discrete energy and radioactivity Solving problems involving atomic spectra PRACTICE: The element helium was first identified by the absorption spectrum of the sun. (a) Explain what is meant by the term absorption spectrum. SOLUTION: An absorption spectrum is produced when a cool gas is between a source having a continuous spectrum and an observer with a spectroscope. The cool gases absorb their signature wavelengths from the continuous spectrum. Where the wavelengths have been absorbed by the gas there will be black lines. continuous spectrum absorption spectrum emission spectrum

16 Topic 7: Atomic, nuclear and particle physics 7.1 Discrete energy and radioactivity Solving problems involving atomic spectra PRACTICE: One of the wavelengths of the absorption spectrum for helium occurs at 588 nm. (b) Show that the energy of a photon having a wavelength of 588 nm is J. SOLUTION: This formula can be used directly: E = hc / λ Where h = Js and is called Planck s constant. energy E of a photon having wavelength λ From E = hc / λ we see that E = ( )( ) / ) E = J.

17 Topic 7: Atomic, nuclear and particle physics 7.1 Discrete energy and radioactivity Solving problems involving atomic spectra PRACTICE: The diagram represents some energy levels of the helium atom. (c) Use the information in the diagram to explain how absorption at 588 nm arises. SOLUTION: We need the difference in energies between two levels to be J. Note that = Since it is an absorption the atom stored the energy by jumping an electron from the J level to the J level, as illustrated.

where n = (an integer) =

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