Question

Integrated Concepts

Estimate the density of a nucleus by calculating the density of a proton, taking it to be a sphere1.2 fm in diameter. Compare your result with the value estimated in this chapter.

Short answer

The density of the proton is mentioned in the chapter as being in the range of 1018 kg/m3. The anticipated density of the proton is 1.1846 times the value in the chapter.

Step-by-step solution

Determine the formulas:

Consider the formula for the density of the proton as follows:

\({\bf{\rho = }}\frac{{\bf{m}}}{{\bf{V}}}\)

Here, m is the mass of the proton and V is the volume.

Step 2:Calculate density of the proton and compare result with the value estimated in this chapter

Given:

The diameter of the sphere is\(d = 1.2\;{\rm{fm}}\).

Finding the density of a proton.

Calculation:

In order to evaluate the density of the proton, we use the following relation:

\(\begin{array}{c}\rho = \frac{{{m_p}}}{{\frac{4}{3}\pi {r^3}}}\\ = \frac{{1.67 \times {{10}^{ - 27}}\;{\rm{kg}}}}{{\frac{4}{3}\pi {{\left( {\frac{{1.2}}{2} \times {{10}^{ - 15}}} \right)}^3}}}\\ = 1.846 \times {10^{18}}\;\frac{{{\rm{kg}}}}{{{{\rm{m}}^3}}}\end{array}\)

As the chapter mention the density of the proton is in range of\({10^{18}}\;\frac{{{\rm{kg}}}}{{{{\rm{m}}^3}}}\).

Therefore, the proton's predicted density is \(1.1846\)times the value in the chapter.