Chapter 32: Nuclear Physics and Nuclear Radiation Selected Solutions

Selected Solutions

7. (a) , where m = (mass of He atom) - (mass of 2 electrons).

(b) The closest approach occurs when all the kinetic energy is converted into electric potential energy.

(c) Since Z = 29 for copper and Z = 79 for gold, the repulsive force the alpha particle experiences is much less when it approaches the copper nucleus. Thus, the distance of closest approach would be less than that found in part (b).

15. (a)

We can use the masses of the atoms since the numbers of electrons will balance

m_i = 211.988842 u
m_f = 207.97664 u + 4.002603 u = 211.97924 u
Dm = m_f - m_i = 211.97924 u - 211.988842 u = -0.00960 u

Therefore,

Again, the atomic masses are

m_i = 239.052158 u
m_f = 235.043925 u + 4.002603 u = 239.046528 u
Dm = m_f - m_i = 239.046528 u - 239.052158 u = -0.005630 u

Therefore,

23. First we need to find the decay constant from the half-life

33. (a) ⁵⁶₂₆Fe

Using the atomic masses,

m_i = 55.934939 u
m_f = 26(1.007825 u) + 30(1.008665 u) = 56.463400 u
Dm = 56.463400 u - 55.934939 u = 0.528461 u

Therefore,

(b) ²³⁸₉₂U

Using the same procedure,

m_i = 238.050786 u
m_f = 92(1.007825 u) + 146(1.008665 u) = 239.984990 u
Dm = 239.984990 u - 238.050786 u = 1.934204 u

Therefore,

51. (a) By definition of the rad:

(b) Using the expression for the amount of heat that produces a specific temperature change,

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