The ratio of mass densities of nuclei of 40Ca and 16O is close to
(a) 5
(b) 2
(c) 0.1
(d) 1

Q: Why do we assume the mass of the nucleus is simply $A \times m_u$?

While there is a slight mass defect due to binding energy, it is extremely small compared to the total mass (usually less than 1%). For calculating bulk properties like density, $A \times m_u$ is a highly accurate and standard approximation.

Q: Is the density of all nuclei exactly the same?

It is approximately the same for all nuclei, except for very light ones (like Hydrogen or Helium) where surface effects are more prominent. For medium and heavy nuclei, the density is remarkably constant.

Q: What is the actual value of this nuclear density?

It is incredibly high, roughly $2.3 \times 10^{17} \text{ kg/m}^3$. To put that in perspective, a single teaspoon of nuclear matter would weigh billions of tons on Earth!

Q: Why does the radius follow the $R = R_0 A^{1/3}$ rule?

This rule arises precisely because nuclear matter is incompressible. If you add more nucleons (increasing $A$), the volume must increase proportionally to accommodate them, which means $R^3 \propto A$, or $R \propto A^{1/3}$.

Q: Does this mean a Calcium nucleus is just a scaled-up Oxygen nucleus?

In terms of density and how tightly the nucleons are packed, yes. However, their internal nuclear structures, energy levels, and overall stability differ significantly due to complex quantum mechanical effects.

Question

The ratio of mass densities of nuclei of $^{40}	ext{Ca}$ and $^{16}	ext{O}$ is close to

Options:

(A) 5

(B) 2

(C) 0.1

(D) 1

Sigma Percentile · Accepted Answer

### The Incompressible Heart of Matter: Understanding Nuclear Density

Imagine you are looking at two very different atomic nuclei: a relatively light Oxygen nucleus ($^{16}	ext{O}$) and a much heavier Calcium nucleus ($^{40}	ext{Ca}$). Calcium has more than double the number of protons and neutrons compared to Oxygen. It is heavier, and it is physically larger.

But what happens if we compare their densities? Does a heavier nucleus pack its protons and neutrons more tightly, or is the spacing between them the same? To answer this, we need to dive into the mathematics of the nucleus.

### The Math of the Nucleus

To find out the density, let's recall the fundamental definition of density. Density ($ho$) is simply the total mass divided by the total volume:

$$ho = \frac{M}{V}$$

For a nucleus, the total mass $M$ is roughly the mass number $A$ (the total number of nucleons) multiplied by the mass of a single nucleon, which we can denote as $m_u$. So, $M \approx A \cdot m_u$.

Assuming the nucleus is roughly spherical, its volume is given by the standard formula for a sphere:

$$V = \frac{4}{3}\pi R^3$$

### The Grand Cancellation

Here is where the physics gets incredibly elegant. The radius of a nucleus $R$ is not arbitrary. Experimental scattering data shows that the radius scales with the mass number according to a very specific relation:

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

where $R_0$ is an empirical constant approximately equal to $1.2 	imes 10^{-15} 	ext{ m}$ (or $1.2 	ext{ fm}$).

If we plug this radius into our volume equation and cube it, the volume becomes:

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

Now, let's bring it all together and substitute the mass and the volume back into the density formula:

$$ho = \frac{A \cdot m_u}{\frac{4}{3}\pi R_0^3 A}$$

Look closely at that expression. We have the mass number $A$ in the numerator and $A$ in the denominator. They perfectly cancel each other out!

$$ho = \frac{m_u}{\frac{4}{3}\pi R_0^3}$$

### The Liquid Drop Analogy

What does this mathematical cancellation actually mean? It means that the density of nuclear matter is completely independent of the mass number $A$. Whether it is a small Oxygen nucleus or a larger Calcium nucleus, the nucleons are packed just as tightly.

Because $m_u$ and $R_0$ are constants, the nuclear density $ho$ is a universal constant for all nuclei, evaluating to an astonishingly high value of roughly $2.3 	imes 10^{17} 	ext{ kg/m}^3$.

Therefore, the ratio of the mass densities of $^{40}	ext{Ca}$ and $^{16}	ext{O}$ is exactly **1:1**. This profound result tells us that nuclear matter is incredibly dense and incompressible, behaving very much like a drop of liquid. This very idea forms the foundational basis of the Liquid Drop Model in nuclear physics!

Animated Solution for Physics - Atoms and Nuclei: The ratio of mass densities of nuclei of $^{40} Ca$ and $^{16} O$ is close to

Select Answer:

Visualized Solution

$^{40} Ca$ and $^{16} O$ Nuclei

Nuclear Density Formula

Radius and Volume

Calculating Density

Conclusion

The Way Forward

The Sigma Insight: Nucleus and Nuclear Reaction

The Incompressible Heart of Matter

The Math of the Nucleus

The Grand Cancellation

The Liquid Drop Analogy

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Animated Solution for Physics - Atoms and Nuclei: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The ratio of mass densities of nuclei of 40Ca and 16O is close to