00:02
So if we start with the equation, khe is equal to m -t -h over m -h -e plus m -t -h times q, that's equal to m -h -e -plus -m -t -h -h -m -t -h -t -h -m -u minus m -t -h -h - minus m -h -e times c squared.
00:37
So plugging what we know into that equation, we have 232 .038 -055 u's, divided by 4 .002603 u's, plus 232 .032 .035 u.
00:55
Times 236 .0456 .056 .5 -6 -8 -8 -ews, u's minus 232 .032 .038 055 use minus 4 .002603us times c squared.
01:20
And this leaves you with 4 .496 m .evs.
01:30
For part b, we use equation 41 -1 to estimate the radii.
01:35
So r of h .e is equal to 1 .2 times 10 to the negative 15 meters times 4 to the 1 third power that leaves you with 1 .905 times 10 to the negative 15 meters for r of t h we have 1 .2 times 10 to the negative 15 meters times 2 .2 to the negative 15 meters times 232 to the third power and that's equal to 7 .374 times 10 to the negative 15 meters for part c the maximum height of the coulon barrier will correspond to the alpha particle and the thorium nucleus being separated by the sum of their radii.
02:23
So u is equal to 1 over 4 pi epsilon not times qhe qth over r -a, and that's equal to 1 over 4 pi -e -e -e -e -t -h -h -e -p -h -r -t -h, over r -h -e plus r -t -h, plugging what we know to that equation, that's 8 .98 times 10 to the 9th, newton's per meter squared divided by c squared, times 2 times 90, times 1 .60 times 10 to the negative 19 c's squared, divided by 4 to the 1 third power plus 232 to the third power times 1 .2 times 10 to the negative 15 meters times 1 .60 times 10 to the negative 13 joules per m .evs...