Atomic and Nuclear Physics (7)

Mr. Klapholz Shaker Heights High School Atomic and Nuclear Physics (7) Problem Solving

Problem 1 Light with a wavelength of 700 nm will appear red. What is the energy (in eV) of one photon of this light?

Solution 1 Energy = Planck’s constant × Frequency E = hf Frequency = ? f = c / l = (3.0 x 108 ) / (700 x 10-9 ) = ? f = 4.3 x 1014 Hz E = (6.6 x 10-34 J s) × (4.3 x 1014 s-1) E = 2.8 x10-19 J E = 2.8 x10-19 J ( 1 eV / 1.6 x10-19 J ) = 1.8 electron Volts

Problem 2 The lowest three energy levels of hydrogen have energies of -13.6 eV, -3.39 eV, and -1.51 eV. • Draw the energy levels. • If the atom had only these three energy levels, then how many different kinds of photons could it emit? • What is the highest-energy photon that the atom can emit?

Solution 2 a) Drawing on the chalkboard. b) 3 energy drops are possible. c) From the highest to the lowest the energy change is: -1.51 eV – (-13.6 eV) = ? = 12.09 eV

Problem 3 Consider uranium-235. • How many protons does it have? • How many neutrons? • What is the symbol for this isotope? • What is the charge of its nucleus, in Coulombs? • What is the approximate mass of the nucleus, in kilograms?

Solution 3 • Every uranium nucleus has 92 protons. • 235 – 92 = 143 neutrons. • 23592U • 92 protons ( 1.602 x 10-19 C / 1 proton ) = ? = 1.47 x 10-17 C e) 235 nucleons ( 1.67 x 10-27 kg / nucleon ) = ? = 3.9 x 10-25 kg

Problem 4 The most common isotope of iron (Fe) has a mass of 53.9396 u. The mass of an isolated proton is 1.00782 u. The mass of a neutron is 1.00866 u. What is the binding energy per nucleon of iron? Find this on the binding energy curve.

Solution 4 Part 1 5426Fe has 26 protons and 28 neutrons. Mass of the protons: 26(1.00782) = ? = 26.203320 u Mass of the neutrons: 28(1.00866) = ? = 28.243480 u Total mass = ? = 54.4458 u Mass Defect = ? = 54.4458 u - 53.9396 u = ? = 0.5062 u

Solution 4 Part 2 1 u = 931.5 MeV (0.5062 u)( 931.5 MeV / 1 u ) = ? = 471.5 MeV Binding energy per nucleon = ? 471.5 MeV / 54 = 8.7 MeV / nucleon Spot this on the curve…

The binding energy curve http://library.thinkquest.org/3471/mass_binding.html

Problem 5 Radium-224 will alpha decay. Write a nuclear equation showing this process.

Solution 5 (part 1)

Solution 5 (part 2) 22488Ra 22088Rn + 42He

Problem 6 How much energy is released by the alpha decay of Radium 224? Masses: Radium-224: 224.0202118 u Radon-220: 220.0113938 u Helium-4: 4.00260325 u

Solution 6 Like a ball rolling downhill, the products have kinetic energy. This energy came from mass! The mass of the products is less than the mass of the reactants. Dm = mass of reactants - mass of products Dm = [ 224.0202118 ] – [ 220.0113938 + 4.00260325 ] Dm = [ 224.0202118 ] – [ 224.013997 ] Dm = 0.006215 u E = mc2 = ( 0.006215 u )( 931.5 MeV / 1 u ) = ? E = 5.789 MeV

Problem 7 Write an equation for the beta decay of carbon-14.

Solution 7 (part 1)

Solution 7 (part 2) 146C 147N + 0-1b + n’ (we can skip the antineutrino) 146C 147N + 0-1e+ n’

Problem 8 Cobalt-60 will beta decay with a half life of about 5 years. If you have a sample of cobalt-60 that is emitting beta particles at the rate of 40 s-1, then what will be the rate in 15 years?

Solution 8 In 5 years the decay rate will be 20 / s. In 10 years the rate will be 10 / s. In 15 years the rate will be 5 / s.

Problem 9 Nitrogen-17 has a half life of about 4 s. If you start with 24 g of nitrogen-14, how much will be left after 16 s?

Solution 9

Problem 10 The half life of carbon-14 is about 5000 years. A sample of wood has 1/16 the amount of carbon-14 that it had when it died. How old is the wood?

Solution 10

Problem 11 A high-energy neutron slams into a nitrogen-14 nucleus. This results in a carbon-14 nucleus, and what else?

Solution 11 146N + 10n 146C + ? 146N + 10n 146C + 11p 146N + 10n 146C + 11H (Any excess energy will go into the KE of the proton)

Problem 12 Use the binding energy curve to determine if it is possible for 236U to split into 92Kr and 141Ba. (The chart is too fuzzy to read accurately, but please try it to see what the task is like.)

Solution 12 (part 1) http://library.thinkquest.org/3471/mass_binding.html

Solution 12 (Part 2)

Solution 12 (Part 3) • The binding energy of a nucleus is the binding energy per nucleon, times the number of nucleons: • BE of 236U: (7 Mev) (236) = 1652 MeV • BE of 92Kr: (8.2 Mev) (92) = 754 MeV • BE of 141Ba: (8 Mev) (141) = 1128 MeV • The binding energy of the products is 1882 MeV. This is greater than the binding energy of the reactants. So…

Solution 12 (Part 4) • The reaction is possible. • The increase in BE / nucleon means that the nuclei are moving toward a more stable state (like a ball rolling downhill). • The increase in BE / nucleon is not like an increase in mass. Mass is like potential energy, so if a reaction decreases the mass, then the reaction can occur…

Can you spot this problem on this graph? http://www.google.com/imgres?imgurl=http://www4.nau.edu/meteorite/Meteorite/Images/BindingEnergy.jpg&imgrefurl=http://www4.nau.edu/meteorite/Meteorite/Book-GlossaryB.html&h=459&w=567&sz=37&tbnid=2l4x3t2tTYh6pM:&tbnh=108&tbnw=134&prev=/images%3Fq%3Dbinding%2Benergy%2Bcurve&zoom=1&q=binding+energy+curve&hl=en&usg=__yHj4cmQEXQXAHu4tw_7fByoPI3s=&sa=X&ei=CXHdTM--Kse1nweDorCDDw&ved=0CCQQ9QEwBA

Tonight’s HW: Go through the Atomic & Nuclear section in your textbook and scrutinize the “Example Questions” and solutions. Bring in your questions to tomorrow’s class.

Atomic and Nuclear Physics (7)