The radius R of a nucleus of mass number A can be estimated by the formula R = (1.3 x 10^-15) A^(1/3

Q: Why do we assume the mass of a proton and a neutron to be exactly equal?

While a neutron is very slightly heavier than a proton, the difference is incredibly small (less than 0.1%). For the purpose of calculating the overall order of magnitude of nuclear density, treating them as equal is a perfectly valid and standard approximation.

Q: Why does the mass number 'A' cancel out in the density formula?

This happens because both the mass and the volume of a nucleus scale linearly with the number of nucleons ($A$). Mass is directly proportional to $A$, and since radius $R \propto A^{1/3}$, the volume ($R^3$) is also directly proportional to $A$. When you divide them, $A$ cancels out.

Q: What does it mean physically that the density is independent of 'A'?

It means that nuclear matter behaves like an incompressible liquid drop. Adding more nucleons to a nucleus just makes the 'drop' bigger, but it doesn't pack the existing nucleons any tighter. This is a key characteristic of the strong nuclear force, known as 'saturation'.

Q: Is the nucleus really a perfect sphere?

Not always. While many light and magic-number nuclei are nearly spherical, many heavier nuclei are deformed (like an American football or a discus). However, for calculating average bulk properties like density, the spherical approximation works exceptionally well.

Q: How can something be so incredibly dense without collapsing into a black hole?

While the density is enormous ($10^{17} \text{ kg/m}^3$), the total mass of a single nucleus is still incredibly tiny. A black hole forms when a massive amount of matter is compressed into a small radius (the Schwarzschild radius). A single nucleus doesn't have nearly enough mass to curve spacetime to that extreme degree.

Question

The radius $R$ of a nucleus of mass number $A$ can be estimated by the formula $R = (1.3 	imes 10^{-15}) A^{1/3}$ m. It follows that the mass density of a nucleus is of the order of ($M_{	ext{proton}} \cong M_{	ext{neutron}} = 1.67 	imes 10^{-27}$ kg)

Options:

(A) $10^{17} 	ext{ kg m}^{-3}$

(B) $10^{24} 	ext{ kg m}^{-3}$

(C) $10^{10} 	ext{ kg m}^{-3}$

(D) $10^{3} 	ext{ kg m}^{-3}$

Sigma Percentile · Accepted Answer

### The Mind-Boggling Density of a Nucleus

Imagine taking the entire mass of a massive battleship and compressing it into a space no larger than a grain of sand. That is the kind of extreme physics we are dealing with when we talk about the nucleus of an atom. In this problem, we are going to calculate the mass density of nuclear matter, and in doing so, we will uncover a beautiful and profound secret about the universe.

### Setting Up the Mass and Volume

To find the density of any object, we need two fundamental ingredients: its total mass and its total volume. Let's start with the mass. A nucleus is made up of protons and neutrons, collectively called nucleons. If a nucleus has a mass number $A$, it means it contains exactly $A$ nucleons. Since the mass of a proton is roughly equal to the mass of a neutron ($m_n \approx 1.67 	imes 10^{-27} 	ext{ kg}$), the total mass $M$ of the nucleus is simply the number of nucleons multiplied by the mass of one nucleon:

$M = A 	imes 1.67 	imes 10^{-27} 	ext{ kg}$

Next, we need the volume. We model the nucleus as a perfect sphere. The volume $V$ of a sphere is given by the classic geometric formula:

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

We are given an empirical formula for the radius of a nucleus: $R = (1.3 	imes 10^{-15}) A^{1/3} 	ext{ m}$. Let's substitute this into our volume equation.

$V = \frac{4}{3}\pi \left(1.3 	imes 10^{-15} A^{1/3}ight)^3$

When we cube the terms inside the parenthesis, something magical happens. The cube of $A^{1/3}$ is simply $A$.

$V = \frac{4}{3}\pi (1.3)^3 	imes 10^{-45} 	imes A 	ext{ m}^3$

### The Beautiful Cancellation

Now, we bring it all together to find the density $ho$, which is mass divided by volume:

$ho = \frac{M}{V} = \frac{A 	imes 1.67 	imes 10^{-27}}{\frac{4}{3}\pi (1.3)^3 	imes 10^{-45} 	imes A}$

Look closely at this equation. The mass number $A$ appears in both the numerator and the denominator. They cancel out completely! This is not just a mathematical convenience; it is a profound physical statement. It tells us that **the density of nuclear matter is completely independent of the size of the nucleus**. Whether it is a tiny Helium nucleus or a massive Uranium nucleus, the nuclear material is packed with the exact same density.

### The Final Calculation

Let's crunch the remaining numbers to find this universal constant:

$ho = \frac{1.67 	imes 10^{-27}}{\frac{4}{3}\pi (1.3)^3 	imes 10^{-45}}$

$ho \approx \frac{1.67}{9.2} 	imes 10^{18} 	ext{ kg m}^{-3}$

$ho \approx 2.3 	imes 10^{17} 	ext{ kg m}^{-3}$

The order of magnitude is clearly **$10^{17} 	ext{ kg m}^{-3}$**. To put this into perspective, a single cubic meter of this material would weigh $10^{17}$ kilograms—that is roughly the mass of a large mountain! This incredible density is a direct result of the strong nuclear force, which binds nucleons together with unimaginable strength.

Animated Solution for Physics - Atoms and Nuclei: The radius $R$ of a nucleus of mass number $A$ can be estimated by the formula $R = (1.3 \times 1 0^{- 15}) A^{1/3}$ m. It follows that the mass density of a nucleus is of the order of ( $M_{proton} ≅ M_{neutron} = 1.67 \times 1 0^{- 27}$ kg)

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Visualized Solution

$Visualizing the Nucleus$

$Mass of the Nucleus$

$Volume of the Nucleus$

$Substituting Radius into Volume$

$Setting up the Density Equation$

$Calculating the Density$

$Final Order of Magnitude$

$The Way Forward$

The Sigma Insight: Nucleus and Nuclear Reaction

The Mind-Boggling Density of a Nucleus

Setting Up the Mass and Volume

The Beautiful Cancellation

The Final Calculation

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Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Visualizing the Nucleus

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Mass of the Nucleus

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Volume of the Nucleus

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Substituting Radius into Volume

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Setting up the Density Equation

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Calculating the Density

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Final Order of Magnitude

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The Way Forward