An alpha (?) particle is the nucleus of a 4 He atom and consists of 2 neutrons a

Question

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

1) What is the energy required to remove the first proton? ?E1 = 19.82 MeV

2) Next, what is the energy required to remove the first neutron? ?E2 = 6.26 MeV

3) Finally, what is the energy required to separate the remaining neutron and proton? ?E3 =

4) What is the total binding energy of the of the ? particle? Ebe =

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

Need help with 3,4,5 please!

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

Explanation / Answer

Answer:

Given data

Masses

31H is 3H of mass 3.01605 u
p is proton of 1.00728 u mass
1u = 931.5 MeV
When a proton is removed. 42Alpha ------> 31H + p

Energy required
=> Ealpha - (E3H + Ep) = 931.5(4.0026 -1.00728-3.01605) = 931.5*(-0.02073) = -19.31 MeV

2.

Energy required to remove neutron
=> E3H -  Emeutron -E2H = 931.5 * (3.01605 - 1.00867 - 2.0141) = 931.5*(-0.00672) = -6.25968 MeV

3.

New species formed 21H ------> p + n

Energy required=

E2H - Ep - En = 931.5 * (2.0141 - 1.00728 - 1.00867) = 931.5 * (-0.00175) = -1.7232 MeV

4) Total binding energy
= Ealpha - 2(Eproton + Eneutron) = 931.5 * (4.0026 - 4.0319) = -27.29295 MeV

5)

Binding energy per nucleon = Ebe / Total no. of nucleons = 27.293/4 = 6.823 MeV

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

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