`.^(238)U=238.05079 u" "._(2)^(4)He=4.00260 u`
`._(90)^(234)Th=234.04363 n" "._(1)^(1)H=1.00783 u`
`._(90)^(237)Pa=237.0512l u`
Here the symbol `Pa` is for the element protacttntum `(Z=91)`.
(b) Show that `.^(238)U`. can not spontaneously emit a proton .

We are given the following atomic masses : ._(92)^(238)U = 238.05079 u " " ._(2)^(4)He=4.00260 u ._(90)^(234)Th=234.04363u " " ._(1)^(1)H=1.00783 u ._(91)^(237)Pa=237.05121 u Here the symbol Pa is for the element protactinium (Z = 91) . Show that ._(92)^(238)U cannot spontaneously emit a proton.

We are given the following atomic masses: ._(92)^(238)U=238.05079u ._(2)^(4)He=4.00260u ._(90)^(234)Th=234.04363u ._(1)^(1)H=1.00783u ._(91)^(237)Pa=237.05121u Here the symbol Pa is for the element protactinium (Z=91)

We are given the following atomic masses: ._(92)^(238)U=238.05079u ._(2)^(4)He=4.00260u ._(90)^(234)Th=234.04363u ._(1)^(1)H=1.00783u ._(91)^(237)Pa=237.05121u Here the symbol Pa is for the element protactinium (Z=91)

We are given the following atomic masses : ._(92)^(238)U = 238.05079 u " " ._(2)^(4)He=4.00260 u ._(90)^(234)Th=234.04363u " " ._(1)^(1)H=1.00783 u ._(91)^(237)Pa=237.05121 u Here the symbol Pa is for the element protactinium (Z = 91) . Calculate the energy released during the alpha decay of ._(92)^(238)U .

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.

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.

By using the following atomic masses : ._(92)^(238)U = 238.05079u . ._(2)^(4)He = 4.00260u, ._(90)^(234)Th = 234.04363u . ._(1)^(1)H = 1.007834, ._(91)^(237)Pa = 237.065121u (i) Calculate the energy released during the alpha- decay of ._(92)^(238)U . (ii) Show that ._(92)^(238)U cannot spontaneously emit a proton.

By using the following atomic masses : ._(92)^(238)U = 238.05079u . ._(2)^(4)He = 4.00260u, ._(90)^(234)Th = 234.04363u . ._(1)^(1)H = 1.007834, ._(91)^(237)Pa = 237.065121u (i) Calculate the energy released during the alpha- decay of ._(92)^(238)U . (ii) Show that ._(92)^(238)U cannot spontaneously emit a proton.