{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

30_InstSolManual_PDF_Part15 - Nuclear and High-Energy...

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
30.60. Set Up: In a time of one half-life the number of radioactive nuclei and the activity decrease by a factor of 2. Note that Solve: (a) 1 half-life: 2 half-lives: 3 half-lives: 4 half-lives: 5 half-lives: (b) In time the activity has decreased by a factor of two n times, so This general result agrees with the results of part (a). (c) so (d) (e) Taking the natural log of both sides of this equation gives and (f) From part (e), 30.61. Set Up: Neglect the mass of the antineutrino. Solve: The mass decrease is The energy equivalent is 30.62. Set Up: The particle emitted in decay is an electron, In a time of one half-life the number of radioactive nuclei decreases by a factor of 2. Solve: (a) The daughter nucleus is (b) 56 yr is so 25% is left. (c) so 30.63. Set Up: The reaction energy is where and If neutral atom masses from Table 30.2 are used, the electron masses cancel. If energy is liberated in the reaction. If energy is absorbed in the reaction. Solve: 10.1 MeV is absorbed.
Background image of page 1
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}