91Ó°ÊÓ

Moment of inertia is defined as the quantity expressed by the body resisting angular acceleration which is the sum of the product of the mass of every particle with its square of a distance from the axis of rotation.

A spinning object with quantized angular momentum is given as follows,

$L^2=I\left(I+1\right)h^2...........\left(1\right)$

Here, L is the angular momentum quantum number, h is reduced Planck’s constant.

Kinetic energy associated with rotational of molecule is given by

$K_{rot}=\frac{L_{rot}^2}{2L}................\left(2\right)$

Here, L is the rotational angular momentum and I is the moment of inertia.

Use equation (1), in equation (2) to find the equation of moment of inertia.

$$\begin{gathered} K_{rot}=\frac{L_{rot}^2}{2I} \\ {K}_{rot}=\frac{I(I+1)h^2}{2I} \\ I=\frac{I(I+1)h^2}{2K_{rot}}............\left(3\right) \end{gathered}$$

Energy of emitted by in form of photon is,

$$\begin{gathered} K_{tot}=87KeV \\ {K}_{tot}=87Kev\left(\frac{{10}^{3}eV}{1KeV}\right)\left(\frac{1.6×{10}^{-19}J}{1eV}\right) \\ {K}_{tot}=1.392×{10}^{-14}J \end{gathered}$$

Substitute (2) for $l,1.05x{10}^{-34}J.s$ for h and $1.392x{10}^{-14}J$for in the equation (3),

$$\begin{gathered} I=\frac{I(I+1)h^2}{2K_{rot}} \\ I=\frac{\left(2\right)\left(2+1\right){\left(1.05×{10}^{-34}J.s\right)}^2}{\left(2\right)\left(1.392×{10}^{-14}J\right)} \\ I=2.37×{10}^{-54}kg.m^2 \end{gathered}$$

Therefore, the moment inertia is $2.37×{10}^{-54}Kg.m^2$.

Let us assume that the shape of the nucleus as spherical, the moment of inertia of a sphere can be calculated as

$I=\frac{2}{5}M{R}^2...........\left(4\right)$

Here, M is the mass of the nucleus and R is the radius of the nucleus.

Substitute $2.37×{10}^{-54}Kg.m^2$ for l and $5.56×{10}^{-27}Kg$for M.

$$\begin{gathered} I=\frac{2}{5}M{R}^2 \\ 2.37×{10}^{-54}=\frac{2}{5}\left(265.56×{10}^{-27}\right)R^2 \\ R=\sqrt{\frac{\left(5\right)\left(2.37×{10}^{-54}\right)}{\left(2\right)\left(\left(265.56×{10}^{-27}\right)\right)}} \\ R=4.72×{10}^{-15}m \end{gathered}$$

Therefore, the radius of the nucleus is $4.72×{10}^{-15}m$.

The radius of a spherical nucleus is given by the following equation,

$$\begin{gathered} R=\left(1.3×{10}^{-15}m\right)A^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.} \\ R=\left(1.3×{10}^{-15}m\right){\left(160\right)}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.} \\ R=7.0×{10}^{-15}m \end{gathered}$$

Therefore, the result in part (b) is 1.5 times the result in part (c).