 ## essay(nuclear/radioactive)

1.(a)State the changes to the nucleus when a nucleus decay by emitting
(i) α-particle
(ii)β-particle
(iii)γ-particle [3 marks]

(b)The success of nuclear reaction using bombardment of α-particle is limited by the energy of
the α-particles produced by the radioactive decay. In an attempt of
producing13N7, protons of
kinetic energy of 2.5 MeV are used to bombard13C6
. Determine quantitatively whether the effort
will be successful. [5 marks]

[Mass of 13C6
= 13.003355u, mass of 13N7 = 13.005739u,
mass of proton = 1.007825u, mass of neutron = 1.0008665u]

(c) In a reaction represented by14N7+4He2->17O8+1H1, nitrogen is bombarded with α-particles
of kinetic energy
7.68 MeV and kinetic energy of oxygen nucleus and proton are 0.56 MeV
and 5.93 MeV respectively. Calculate the mass of the α-particle in u, the unified atomic mass.
[7 marks]

[Mass of 14N7
= 14.003074u, mass of 17O8= 16.999134u,
mass of 1H1= 1.007825u, mass of 1n0= 1.0008665u]

(a)When a nucleus decays by emitting
i) α-particle
The charge of is conserved because proton number is conserved. The nucleon number is
also conserved.

ii) β-particle

The number of neutron decreases by 1, the number of proton increases by 1, and the
nucleon number remains unchanged.

iii)γ-particle

It does not result in any changes to the proton number and nucleon number because γ-
particle is a photon (electromagnetic radiation without any charge)

b) 13C6+1H1->13O7+1n0

The total mass of 13C6
and 1H1= (13.003355 + 1.007825) u= 14.01118 u

The total mass of 1307and 1n0= (13.005739 + 1.008665) u= 14.014404 u

The total mass before reaction is smaller than that after the reaction, the reaction does not
occur.

c)Δm c2 = 7.58 – (0.56 + 5.93) MeV= 1.19 MeV
Δm= 1.19 MeV/(3x10^8)2=2.11843447 × 10-30 kg = 0.001276165 u

Δm= ( mN + mα) – (mO + mp)

0.001276165 u= (14.003074 + mα) u – ( 16.999134 + 1.007825) u

mα = 0.001276165 - 14.003074 +16.999134 + 1.007825
= 4.00516 u Giraffe  1018  ## 回复： essay(nuclear/radioactive) Joker 5  