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PROBLEMS

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

2.

a. Given the following masses, 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?

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

ANSWERS

1.

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

2.

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:

Conservation of momentum gives us:

m_Thv_Th+ m_Hev_He = 0

and

v_Th+ = -(m_He/m_Th)v_He [eqn. 1]

The energy released is in the form of kinetic energy:

(1/2)m_Thv_Th² + (1/2)m_Hev_He² = E_k

Substituting eqn. 1 and simplifying:

v_He = √{[2E_k] / [(m_He)(1 + m_He/m_Th)]} = 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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