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**Hint:**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

$d = \dfrac{m}{V}$

Where

m $ = $ mass of nucleus

V $ = $ volume of nucleus

**Complete step by step answer:**

(a) Mass of tangent nucleus ${(_1}{H^2}){m_t} = 2.0147amu$

Mass of bombarding nucleus ${(_1}{H^2}){m_b} = 2.0147amu$

Mass of product nucleus ${(_2}H{e^4}){m_\rho } = 3.0169amu$

Mass of outgoing particle ${(_0}{n^1}){m_n} = 1.0087amu$

So, ${m_t} + {m_b} = 2.0147 + 2.0147$

$\Rightarrow {m_t} + {m_b} = 4.0294amu$

$\Rightarrow {m_\rho } + {m_n} = 3.0169 + 1.0087$

$\Rightarrow {m_\rho } + {m_n} = 4.0256amu$

Hence $({m_t} + {m_b}) - ({m_\rho } + {m_n}) = 0.0038amu$

i.e.., $({m_t} + {m_b}) > ({m_\rho } + {m_n})$

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

$\Rightarrow R = {R_0}{A^{1/3}}$

Where

$\Rightarrow {R_0} = $ Fermi radius

A $ = $ Mass number

So, the volume of nucleus

$\Rightarrow V = \dfrac{4}{3}\pi {R^3}$

$\Rightarrow V = \dfrac{4}{3}\pi {({R_0}{A^{1/3}})^3}$

$\Rightarrow V = \dfrac{4}{3}\pi R_0^3A$

$\Rightarrow V \propto A$ …..(1)

Nuclear density $ = \dfrac{{Mass}}{{Volume}} = \dfrac{m}{V}$

Mass can be written as

$m = A$ $($Mass number$)$

Nuclear density $ = \dfrac{A}{{\dfrac{4}{3}\pi R_0^3A}}$

Nuclear density $ = \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, $_1^1H,_1^2H$ and $_1^3H$.

The first two of these are stable while $_1^3H$ has a half life of $12.32$ years.

There are also heavier isotopes, which are all synthetic and have a half life less than ${10^{ - 21}}$ second of these $_1^5H$ is the most stable and $_1^7H$ is the least.

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