Q2. Using the Semi Emperical Mass Formula, Determine a) The...

Q2. Using the Semi Emperical Mass Formula, Determine a) The neutron and proton separation energies for a nucleus $^{A}X_{Z}$. b) The Q- value for $\alpha$ decay of a nucleus $^{A}X_{Z}$. c) The Q- value of a fusion reaction in which identical nuclei fuse to form the nucleus $^{A}X_{Z}$. d) The Q- value of a fission reaction in which two identical daughters are emitted from a single nucleus $^{A}X_{Z}$. Note A and Z must both be large enough for the Q to be positive. You may replace Z(Z-1) by its approximation $Z^{2}$. Ignore the pairing terms.

Transcript

00:01 So given in the question table presenting the mass axis in units of kilo electron volts per c square for several nuclides is given as follows.

00:23 This is the table given and it has been asked to find in part a the atomic mass in atomic mass units u for each of the nuclides or elements in the table.

00:46 In part b it has been asked to find the mass energy difference between the initial and the final state for the following fission reactions and the reactions are 252 cf 98 breaks into undergo fission into barium molybdenum and three neutrons barium mass number is 140 mo mass number is 109 and neutron mass number is 1 and 256 fm undergo fission to 140 xe 112 pd and four neutrons and in part c you have to find out whether these new these fission reactions occur spontaneously or not.

01:59 So solving the first part we know the mass axis that delta m is basically the atomic mass in units of atomic mass unit u minus the mass number that is capital a.

02:16 Therefore the atomic mass u would be delta m plus a where delta m would be in units of u and we know the energy equivalent of one u is 931 .49 million electron volts so we can write one u as 931 .49 million electron volts per c square for neutron it has been already given so coming to the second element that is cf 252 so delta m would be delta m given is 76034 kilo electron volt per c square we divide it with 931 .49 into 10 to the power 3 because 931 .49 is a million electron volts so convert it into kilo electron volts we get this value to be 0 .08162621 times u and if we add it with the mass number we get the atomic mass and that comes out to be 252 .0816262 u.

03:30 In the same fashion we can find out the atomic mass of all other elements provided in the table for example for fm 256 delta m given is 85496 kilo electron volt per c square which comes out to be 0 .09178413 times 43413 u so the atomic mass comes out to be 256 .09178413 u.

04:12 Similarly for 140 barium ba delta m is given to be negative minus 83271 kilo electron volt per c square so this delta m value will be subtracted from the mass number to find the atomic mass this comes out to be minus 0 .08939548 times u so it is atomic mass is 140 minus 0 .088939548 u that is 139 .91060452 u.

04:57 So coming to 140 xe delta m is given to be again negative and that value is minus 72990 kilo electron volt per c square which is minus 0 .07835833 u so atomic mass would be 140 minus this value which comes out to be 139 .92164167 u.

05:40 Coming to palladium 112 pd 112 delta m is given to be 86336 kilo electron volt per c square which is equal to minus 0 .09268591 u so atomic mass would be 112 minus this value which comes out to be 111 .90731409 u and the seventh one is of molybdenum mo 109 so delta m given is minus 0 minus 67250 kilo electron volt per c square which is minus 0 .07219616 u so the atomic mass is 109 minus this value and which comes out to be 108 .92780384 u.

06:54 So the table with the atomic mass or representing the atomic mass of the given nucleoids or elements are as follows.

07:16 So we can write the values of the atomic mass as was calculated in the above calculations so it is this one these are the values for all the elements given in the table.

07:47 Finally this mo comes out to be 108 .9278038 u...

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