Atomic and Nuclear Physics (7)

Problem Solving

Mr. KlapholzShaker Heights

High School

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

Solution 1Energy = Planck’s constant × Frequency

E = hfFrequency = ?

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

Solution 2a)

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 3Consider uranium-235.

a) How many protons does it have?b) How many neutrons?c) What is the symbol for this isotope?d) What is the charge of its nucleus, in Coulombs?e) What is the approximate mass of the nucleus, in kilograms?

Solution 3a) Every uranium nucleus has 92 protons.b) 235 – 92 = 143 neutrons.c) 235

92U

d) 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 4The 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 154

26Fe 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 uTotal mass = ?= 54.4458 u

Mass Defect = ?= 54.4458 u - 53.9396 u = ?

= 0.5062 u

Solution 4 Part 21 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 5Radium-224 will alpha decay.

Write a nuclear equation showing this process.

Solution 5 (part 1)

Solution 5 (part 2)224

88Ra 22088Rn + 4

2He

Problem 6How much energy is released by the alpha

decay of Radium 224?Masses:

Radium-224: 224.0202118 uRadon-220: 220.0113938 u

Helium-4: 4.00260325 u

Solution 6Like 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 7Write an equation for the beta decay of carbon-14.

Solution 7 (part 1)

Solution 7 (part 2)14

6C 147N + 0

-1 b + n’

(we can skip the antineutrino)14

6C 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 9Time / s Nitrogen remaining / g

0 244 128 6

12 316 1.5

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

Time / years C-14 remaining / g0 16

5000 810000 415000 220000 1

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 + 1

0n 146C + ?

146N + 1

0n 146C + 1

1p 14

6N + 10n 14

6C + 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)

Isotope Binding Energy per nucleon / MeV

uranium-236 7

krypton-92 8.2

barium-141 8

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.