Saved Bookmarks
| 1. |
We are given the following atomic masses: ""_(92)^(238) U = 238.05079 u "" _(2)^(4)He = 4.00260 u ""_(90)^(234)Th = 234.04363 u""_(1)^(1)H= 1.00783 u ""_(91)^(237)Pa = 237.05121 u Here the symbol Pa is for the element protactinium (Z = 91). (a) Calculate the energy released during the alpha decay of ""_(92)^(238)U. (b) Show that ""_(92)^(238)U can not spontaneously emit a proton. |
|
Answer» Solution :(a) The alpha decay of `""_(92)^(238)U` is GIVEN by Eq.The energy released in this process is given by `Q = (m_u - m_(He))c^2` Substituting the atomic masses as given in the data. We find `Q = (238.05079 - 234.04363 - 4.00260) u xx c^2` `= (0.00456 u)c^2` ` = (0.00456 u)(931.5 MEV//u)` `= 4.25 MeV`. (b) If `""_(92)^(238)U` SPONTANEOUSLY emitts a PROTON, the decay process would be `""_(92)^(238)U to ""_(91)^(237)U + ""_(1)^(1)H` The Q for this process to happen is `= (m_U - m_(Pa) - m_(H)) c^2` `=(238.05079 - 237.05121 - 1.00783) u xx c^2` `= (-0.00825 u)c^2` `= -(0.00825 u)(931.5 MeV//u)` `= -7.68 MeV`. Thus, the Q of the process is negative and therefore it cannot proceed spontaneously. We will have to supply an energy of `7.86 MeV` to a `""_(92)^(238)U` nucleus to make it emit a proton. |
|