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帖子 由 Giraffe 于 周六 七月 25, 2009 12:45 pm

1.(a)State the changes to the nucleus when a nucleus decay by emitting
(i) α-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.


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

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

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回复: essay(nuclear/radioactive)

帖子 由 Joker 于 周二 十月 11, 2011 10:38 am

加油加油!!! Smile

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注册日期 : 11-10-11


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