ch42-p006 - (d) In the unit MeV/ c 2 , = (1.008665 u...

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

View Full Document Right Arrow Icon
6. (a) Table 42-1 gives the atomic mass of 1 H as m = 1.007825 u. Therefore, the mass excess for 1 H is = (1.007825 u – 1.000000 u)= 0.007825 u. (b) In the unit MeV/ c 2 , = (1.007825 u – 1.000000 u)(931.5 MeV/ c 2 ·u) = +7.290 MeV/ c 2 . (c) The mass of the neutron is given in Sample Problem 42-3. Thus, for the neutron, = (1.008665 u – 1.000000 u) = 0.008665 u.
Background image of page 1
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: (d) In the unit MeV/ c 2 , = (1.008665 u 1.000000 u)(931.5 MeV/ c 2 u) = +8.071 MeV/ c 2 . (e) Appealing again to Table 42-1, we obtain, for 120 Sn, = (119.902199 u 120.000000 u) = 0.09780 u. (f) In the unit MeV/ c 2 , = (119.902199 u 120.000000 u) (931.5 MeV/ c 2 u) = 91.10 MeV/ c 2 ....
View Full Document

Ask a homework question - tutors are online