The nucleus dysprosium-160 (containing 160 nucleons) acts like a spinning object with quantized Angular momentum. $L^2=l\left(I+1\right)h^2$, and for this nucleus it turns out thatmust be an even integer . When a Dy-160 nucleus drops from the l = 2 state to the l = 0 state, it emits an 87 KeV photon . (a) what is the moment of inertia of the Dy-160 nucleus? (b) Given your result from part (a), find the approximate radius of the Dy-160 nucleus, assuming it is spherical. (In fact, these and similar experimental observation have shown that some nuclei are not quite spherical.) (c) The radius of a (spherical) nucleus is given approximately by$(1.3x{10}^{-15}\mathrm{m})A^{\frac{1}{3}}$ , where A is the total number of protons and neutrons. Compare this prediction with your result in part (b).

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,

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

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

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

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

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

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

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

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

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

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