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Nuclear Physics and Radiation

Nuclear Physics and Radiation Problems

Solutions are printed below.

1. Given the masses in this table, find the mass defect and the binding energy of a helium-4 nucleus.

particlemass (kg)
helium-4 nucleus 6.6447X10-27

2. a. Given the masses in this table, find the energy released when U-238 transmutes to Th-234 by alpha decay.
b. Ignoring relativistic effects, what is the speed of the alpha particle as a result of the alpha decay?

particle mass (amu)
U-238 238.0508
Th-234 234.0436
He-4 4.0026

3.What is the wavelength of a 0.217 MeV photon emitted during gamma decay?

Answers to Nuclear Physics and Radiation Problems

a. sum of masses of individual nucleons = 2(1.6726X10-27) + 2(1.6749X10-27) = 6.6950X10-27 kg
mass defect = 6.6950X10-27 - 6.6447X10-27 = 0.0503X10-27 kg

b. binding energy = (Δm)c2 = (0.0503X10-27)(3.00X108) = 4.53X10-12 J
(4.53X10-12 J)/(1.60X10-19) = 2.83X107 eV = 28.3 MeV

a. mass defect: 238.0508 - (234.0436 + 4.0026) = 0.0046 u
energy equivalent: (0.0046 u)(931.5 MeV/u) = 4.3 MeV

b. The values for part a are converted to mks units:

quantitysymbolconversion factor value
energy released Ek 1.602x10-19 J/eV 6.88886x10-13 J
mass of Th-234 mTh 1.66053886x10-27 kg/amu 3.8864x10-25 kg
mass of He-4 mHe 1.66053886x10-27 kg/amu 6.6465x10-27 kg
speed of Th-234 vTh - ?
speed of He-4 vHe - ?

Conservation of momentum gives us:
mThvTh+ mHevHe = 0
vTh+ = -(mHe/mTh)vHe [eqn. 1]

The energy released is in the form of kinetic energy:
(1/2)mThvTh2 + (1/2)mHevHe2 = Ek

Substituting eqn. 1 and simplifying:
vHe = √{[2Ek] / [(mHe)(1 + mHe/mTh)]} = 1.4x107 m/s

3. The energy is converted to Joules:
(0.217x106 eV) (1.602x10-19 J/eV) = 3.48x10-14 J

The energy is converted to wavelength:
λ = hc/ΔE = (6.63x10-34 J*s)(3.00x108 m/s)/(3.48x10-14 J) = 5.72x10-12 m

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