Question

An alpha (α) particle is the nucleus of a ⁴He atom and consists of 2 neutrons and two protons bound together. Let's take apart an α particle, step by step, looking at the energy required at each step. To do so, we may want to use the following atomic masses:

4He	4.0026 u		2H	2.0141 u
3H	3.01605 u		2H	1.00783 u
p	1.00728 u		n	1.00867 u
e	5.48 × 10-4 u

1)What is the energy required to remove the first proton?

ΔE₁ =

2) Next, what is the energy required to remove the first neutron?

ΔE₂ =

3) Finally, what is the energy required to separate the remaining neutron and proton?

ΔE₃ =

4) What is the total binding energy of the of the α particle?

E_be =

5) What is the total bidning energy per nucleon for the α particle?

E_ben =

Answer 1

Answer

e2 na 85 - Concept - initialmass.) Energy required to lemove the nucleons 1) mas main 4tte = 4.00260 mass of 3H tip - 3.016050 +1.00728u = 4.02333 0 mf = Ama mf mi 4.0 2333 – 4.0096u = 0.020730 ofis 6. 020 73 ) x 9315 = 19.309995 MeV= 19.31 MeV. mass of 34 = 3.0160su. mf mass of 2H+ In= 2.0thl ut 1:008670=3.0 22770 om a me- mi - 3.022770 - 3.016050 = 0.00672 u. (0.00672) x 931.5 = 6.25968 Meu = 6.26 Mev o Era mf z ma onans of 24 = 2.01410 marr z 24 (afta sonoval) + Ipt in 1.007830 + 1.00728 ut 1.00867 3.023780 om = mf-mi = 3.023780 2-01410F 1.009680 AF 1.00968 x 93105 = 940.51692 MeV a 940.52 Mev mass 4) To find Binding energy maso a proto ton = 2x1-00728= 2.01456U Marss of 2 neutrow = 2x 1.00867= 2. 01734u Total 4.031900 Marss of 4te 4.00 260 u Masss defect 0.02930u. Bruding enasy Ebe -0.0293)x 937.5 2 27.29295 = 27.293 Mev 27.293 6.82 Mev. Eben --