Question

Uranium-238 ( ${\mathbf{U}}^{\mathbf{238}}$) has three more neutrons than uranium-235 ( ${\mathbf{U}}^{\mathbf{235}}$). Compared to the speed of sound in a bar of ${\mathbf{U}}^{\mathbf{235}}$, is the speed of sound in a bar of ${\mathbf{U}}^{\mathbf{238}}$ higher, lower, or the same? Explain your choice, including justification for assumptions you make.

Short answer

Speed of sound in ${\mathrm{U}}^{238}$ is lower than speed of sound in ${\mathrm{U}}^{235}$ because speed of sound is inversely proportional to the root of atomic mass.

Step-by-step solution

Given data

Bars of ${\mathrm{U}}^{235}$ and ${\mathrm{U}}^{238}$.

It is given that the uranium ${\mathrm{U}}^{238}$has three more neutrons as compared to uranium ${\mathrm{U}}^{235}$.

Speed of sound in solid

Speed of sound in a solid of stiffness constant $k$, atomic mass $m$and interatomic bond length $d$ is given by,

$\mathbf{v}\mathbf{=}\sqrt{\frac{\mathbf{k}}{\mathbf{m}}}\mathbf{d}\mathbf{}....\left(\mathrm{i}\right)$

Comparing speed of sound in U238 to that in U235

${\mathrm{U}}^{238}$has three more neutrons in its nucleus than ${\mathrm{U}}^{235}$. Thus the atomic mass of ${\mathrm{U}}^{238}$is greater than that of ${\mathrm{U}}^{235}$. Their stiffness constant and interatomic distance are the same. From equation (i), it is seen that the speed of sound in a solid is inversely proportional to the root of its atomic mass. Thus a solid with greater atomic mass will have lower speed of sound in it. Hence speed of sound in ${\mathrm{U}}^{238}$is lower than that of ${\mathrm{U}}^{235}$