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: 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

Answer 1

1)

When a proton is removed, following reaction occurs:

⁴₂Alpha ------> ³₁H + p

³₁H is 3H of mass 3.01605 u
p is proton of 1.00728 u mass
1u = 931.5 MeV

Energy required=
E_alpha - (E_3H + E_p)
= 931.5(4.0026 -1.00728-3.01605)
= 931.5*(-0.02073)
= -19.31 MeV (negative sign shows energy absorbed)

So, E1 = 19.31 MeV

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2)

After removal of neutron, new species formed is ²₁H
Energy required to remove neutron
=E_3H -  E_meutron -E_2H
= 931.5 * (3.01605 - 1.00867 - 2.0141)
= 931.5*(-0.00672)
= -6.25968 MeV

So, E2 = 6.26 MeV

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3)

New species formed
²₁H ------> p + n

Energy required=

E_2H - E_p - E_n
= 931.5 * (2.0141 - 1.00728 - 1.00867)
= 931.5 * (-0.00175)
= -1.7232 MeV

E3 = 1.7232 MeV

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4) Total binding energy
= E_alpha - 2(E_proton + E_neutron)

= 931.5 * (4.0026 - 4.0319)
= -27.29295 MeV

Ebe is 27.293 MeV

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5) Binding energy per nucleon
= Ebe/total number of nucleons

= 27.293/4

= 6.823 MeV

So Eben = 6.823 MeV