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Question: (a) In a typical nuclear reaction e.g. \(_1^2H + _1^2H \to _2^3He + _1^2n + 3.28 MeV\) Although ...

(a) In a typical nuclear reaction e.g.
12H+12H23He+12n+3.28MeV_1^2H + _1^2H \to _2^3He + _1^2n + 3.28 MeV
Although a number of nucleons is conserved, yet energy is released, How ? Explain.
(b) Show that nuclear density in a given nucleus is independent of mass number A.

Explanation

Solution

In order to solve part (a) calculate the total mass of nuclei before reaction and total mass of product nuclei and nucleons and compare them.
(b) For part (b) first calculate radius of nucleus and volume of nucleus and put in formula of nuclear density which is
d=mVd = \dfrac{m}{V}
Where
m == mass of nucleus
V == volume of nucleus

Complete step by step answer:
(a) Mass of tangent nucleus (1H2)mt=2.0147amu{(_1}{H^2}){m_t} = 2.0147amu
Mass of bombarding nucleus (1H2)mb=2.0147amu{(_1}{H^2}){m_b} = 2.0147amu
Mass of product nucleus (2He4)mρ=3.0169amu{(_2}H{e^4}){m_\rho } = 3.0169amu
Mass of outgoing particle (0n1)mn=1.0087amu{(_0}{n^1}){m_n} = 1.0087amu
So, mt+mb=2.0147+2.0147{m_t} + {m_b} = 2.0147 + 2.0147
mt+mb=4.0294amu\Rightarrow {m_t} + {m_b} = 4.0294amu
mρ+mn=3.0169+1.0087\Rightarrow {m_\rho } + {m_n} = 3.0169 + 1.0087
mρ+mn=4.0256amu\Rightarrow {m_\rho } + {m_n} = 4.0256amu
Hence (mt+mb)(mρ+mn)=0.0038amu({m_t} + {m_b}) - ({m_\rho } + {m_n}) = 0.0038amu
i.e.., (mt+mb)>(mρ+mn)({m_t} + {m_b}) > ({m_\rho } + {m_n})
So, this extra mass is counter into energy and this energy is released although a number of nucleons are consumed.
(b) The radius of nucleus is given as
R=R0A1/3\Rightarrow R = {R_0}{A^{1/3}}
Where
R0=\Rightarrow {R_0} = Fermi radius
A == Mass number
So, the volume of nucleus
V=43πR3\Rightarrow V = \dfrac{4}{3}\pi {R^3}
V=43π(R0A1/3)3\Rightarrow V = \dfrac{4}{3}\pi {({R_0}{A^{1/3}})^3}
V=43πR03A\Rightarrow V = \dfrac{4}{3}\pi R_0^3A
VA\Rightarrow V \propto A …..(1)
Nuclear density =MassVolume=mV = \dfrac{{Mass}}{{Volume}} = \dfrac{m}{V}
Mass can be written as
m=Am = A ((Mass number))
Nuclear density =A43πR03A = \dfrac{A}{{\dfrac{4}{3}\pi R_0^3A}}
Nuclear density =34πR03 = \dfrac{3}{{4\pi R_0^3}}

Above expression shows that nuclear density is independent of mass number A.

Note: Many times, students may get confused between isotopes of Hydrogen. So, always remember that hydrogen has three naturally occurring isotopes, 11H,12H_1^1H,_1^2H and 13H_1^3H.
The first two of these are stable while 13H_1^3H has a half life of 12.3212.32 years.
There are also heavier isotopes, which are all synthetic and have a half life less than 1021{10^{ - 21}} second of these 15H_1^5H is the most stable and 17H_1^7H is the least.