Physics 10262 - Chapter 1 – Homework

1. Calculate the wavelength of a: γ-ray photon with E=1.46 MeV

x-ray photon with E=54.0 keV UV photon with E =10.6 eV Visible light with E = 2.2 eV Infrared light with E = 0.1 eV Radio waves with E = 10-7 eV (1 eV = 1.6·10-19 J)

Solution:

γ-ray photon with E=1.46 MeV λ=8.49·10-13 m

x-ray photon with E=54.0 keV λ=2.29·10-11 m UV photon with E =10.6 eV λ=1.17·10-7 m Visible light with E = 2.2 eV λ=5.64·10-7 m Infrared light with E = 0.1 eV λ=1.24·10-5 m Radio waves with E = 10-7 eV λ=1.24·10+1 m

2. Calculate the energies and frequencies of the photons of the Balmer series in a

hydrogen atom up to n=4. Solution:

n=3: E=1.89 eV, v=4.57·1014 [1/s] n=4: E=2.55 eV, v=6.17·1014 [1/s]

3. Calculate the wavelength and the energy of photon emitted from transitions between the L-shell (n=2) and the K-shell (n=1) for hydrogen (Z=1), carbon (Z=6), Fe (Z=26), copper, Z=29, silver (Z=47), gold (Z=79), and lead (Z=82).

Solution:

Hydrogen: 10.20 eV; 1.22·10-7 m Carbon: 255 eV; 4.86·10-9 m Iron: 6.38 keV; 1.95·10-10 m Copper: 8.0 keV; 1.55·10-10 m Silver: 21.58 keV; 5.75·10-11 m Gold: 62.06 keV; 1.99·10-11 m Lead: 66.92 keV; 1.85·10-11 m

4. The Oxygen isotope 18O has a mass number A=18, what is its number of neutrons?

The Oxygen isotope 13O cannot maintain its stability because of the deflective Coulomb forces between its positively charged protons, what is the neutron number? 209Bi (bismuth) is the heaviest stable element with one more proton than Lead, determine it’s neutron number. 197Au is the only stable Gold isotope, what is its number of neutrons?

Solution:

18O: A=18, Z=8, N=10 13O: A=13, Z=8, N=5 209Bo: A=209, Z=83, N=126 197Au: A=197, Z=79, N=118

5. The radioactive fluorine isotope 18F decays by what mechanism to 18O (oxygen)?

The radioactive gold isotope 198Au decays by what mechanism to 198Hg (mercury)? The radioactive radium isotope 226Ra decays by what mechanism to 222Rn (radon)?

Solution:

18F⇒18O: Z=9⇒Z=8: β+-decay

198Au⇒198Hg: Z=79⇒Z=80: β--decay 226Ra⇒222Rn: A=226⇒A=222: α-decay

6. The aluminum isotope 26Al decays by β+ decay to ….?

The gadolinium isotope 148Gd decays by β+ decay to .…? The carbon isotope 14C decays by β- decay to ….?

Solution: The aluminum isotope 26Al decays by β+ decay to 26Mg? The gadolinium isotope 148Gd decays by β+ decay to 144Sm? The carbon isotope 14C decays by β- decay to 14N?

7. Calculate the decay constant λ for the: Bismuth isotope 108Bi with a half life of 3.7·105 y; Potassium isotope 40K with a half life of 1.28·109 y; Carbon isotope 14C with a half life of 5.73·103 y; Cobalt isotope 60Co with a half life of 5.27 y.

Solution:

108Bi with a half life of 3.7·105 y: λ = 1.85·10-6 y-1 40K with a half life of 1.28·109 y: λ = 5.42·10-10 y-1 14C with a half life of 5.73·103 y: λ = 1.21·10-4 y-1

60Co with a half life of 5.27 y: λ = 1.32·10-1 y-1

8. The body of Ramses II contained 3.4·1020 particles of the Potassium isotope 40K and 1.6·1015 particles of the radioactive Carbon 14C at the moment of his death. How many 40K and 14C particles are left in the mummy after 3290 years in the tomb?

Solution:

9. Calculate the present 14C and 40K activity of the mummy of Ramses II. Solution:

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A = λ ⋅ N;A14C

= λ ⋅1.075 ⋅1015 =1.21⋅10−4 ⋅1.075 ⋅1015 =1.3 ⋅1011y−1

A 40K= λ ⋅ 3.4 ⋅1020 = 5.42 ⋅10−10 ⋅ 3.4 ⋅1020 =1.8 ⋅1011y−1

10. During the 3290 years in the tomb, the mummy of

Ramses was exposed to a cosmic ray bombardment of 1·104 particles/m2s. This radiation deposits at average an energy of ΔE/Δx = 3·104eV/m into the body, calculate the total dose the mummy received since the burial feasts! Assume an average mummy material density of ρ = 1.2·103 kg/m3. Calculate the total dose!

Solution:

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D =EM

= 3 ⋅10−4 MeVcm

⋅1⋅104 particles

m2s

⋅

t y[ ]ρ kg /m3[ ]

D = 3 ⋅10−4 MeVcm

⋅1 particles

cm2s

⋅3290 ⋅ 3.15 ⋅107 s[ ]1.2 ⋅10−3 kg /cm3[ ]

= 2.6 ⋅1010 MeVkg

D = 2.6 ⋅1010 MeVkg

= 4.16 ⋅10−3 Gy[ ]