# Nuclear Radius

## Nuclear Radius Revision

**Nuclear Radius**

As the **nucleus** of an atom is incredibly small, its radius cannot be directly measured. Instead, scientists have come up with methods for estimating the **nuclear radius** to high levels of **accuracy**.

**Distance of Closest Approach Method**

As seen previously, if a beam of **alpha particles** are fired at a sheet of **gold foil**, some of the **alpha particles** will **reflect** back to the source. The reason for this phenomenon is the **electrostatic repulsion** between the **positive alpha particle** and the **positive nucleus**.

At the point of closest approach, the **maximum kinetic energy** of the alpha particle must **equal** the **maximum electric potential energy**. Therefore:

Hence:

\dfrac{1}{2}mv^2 = \dfrac{Qq}{4 \pi \epsilon_0 r}r = \dfrac{Qq}{2 \pi \epsilon_0 m v^2}

- Q= the
**charge of the nucleus**in coulombs \text{(C)} - q= the
**charge of an alpha particle**in coulombs (= 3.2 \times 10^{-19} \text{ C}) - \epsilon_0= the
**permittivity of free space**(= 8.85 \times 10^{-12} \text{s}^4 \text{A}^2 \text{m}^{-3} \text{kg}^{-1}) - m= the
**mass of an alpha particle**in kilograms \text{(kg)} - v= the
**velocity of the alpha particle**in metres per second (\text{ms}^{-1})

Although this method gives an accurate value for the distance of closest approach, it **does not give the exact radius** of the nucleus. This is because the value includes the **distance** between the alpha particle and the nucleus which do not touch. Therefore the value will always be an **overestimate**.

Some typical values for **nuclear radius** are shown in the table below:

Element |
Nuclear Radius (m) |

Carbon | 2.66 \times 10^{-15} |

Iron | 4.35 \times 10^{-15} |

Lead | 6.66 \times 10^{-15} |

**Electron Diffraction Method**

If a beam of **electrons** are fired at a **diffraction grating** at speeds close to the speed of light, they act like a wave and diffract with a **de Broglie wavelength** equal to:

\lambda = \dfrac{h}{mv}

- \lambda= the
**de Broglie wavelength**in metres \text{(m)} - h=
**Planck’s constant**(= 6.63 \times 10^{-34} \text{ kgm}^2\text{s}^{-1}) - m= the
**mass of an electron**(=9.11 \times 10^{-31} \text{ kg}) - v= the
**velocity of the electron**in metres per second \text{(ms}^{-1})

If the **electron** is passed through the **diffraction grating** it forms a pattern of concentric rings made of **minima** and **maxima** on a screen. By measuring the angle to the **first minima**, the **radius of a nucleus** can be calculated using:

R= 1.22 \dfrac{\lambda}{2 Sin \theta}

- \lambda= the
**de Broglie wavelength**in metres \text{(m)} - \theta = the
**angle to the first minimum**in degrees or radians \degree \text{ or rad})

This method gives a very accurate, **direct measurement for the radius of the nucleus**. However, for this method to work the **electrons** must be close to the speed of light which can be difficult to achieve.

A graph of **diffraction angle** against **intensity of electrons** can be plotted:

The **first minima** can be found as the minima closest to the **central maxima**. On a graph of intensity against angle, the angle can be determined and placed in the equation above to calculate **nuclear radius**.

**Radius and Nucleon Number**

Through repeated experimentation with different atoms, an equation for calculating the **nuclear radius** was formed.

R=\dfrac{1}{3} R_0 A

- R= the
**nuclear radius**in metres \text{(m)} - R_0= a
**constant of proportionality**(=1.05 \text{ fm}) - A= the
**mass number**of the atom

**Example:** Calculate the nuclear radius of carbon-12.

**[2 marks]**

**Nuclear Density**

Assuming the **nucleus is a sphere**, it’s **volume** can be calculated using the equation:

V=\dfrac{4}{3} \pi R^3

- V=
**volume**in cubic metres \text{(m}^3) - R= the
**nuclear radius**in metres \text{(m)}

We have just learnt that the **nuclear radius **can be calculated using R=\dfrac{1}{3}R_0 A. Sunstituting this into the above equation gives:

As **density** is the **mass per unit of volume **and:

\rho = Au

- \rho=
**density**in kilograms per cubic metre \text{kgm}^{-1} - A= the
**atomic number** - u= the
**atomic mass unit**(=1.66 \times 10^{-27} \text{ kg})

Therefore:

\begin{aligned} \rho &=\dfrac{Au}{\dfrac{4}{3}\pi R_0^3 A} \\ \rho &= \dfrac{3u}{4 \pi R_0^3} \end{aligned}

## Nuclear Radius Example Questions

**Question 1:** What are the benefits of using the electron diffraction method to measure the nuclear radius?

**[2 marks]**

You would obtain a **direct measure for the radius** rather than an **overestimate produced by the distance of closest approach**.

**Question 2:** An electron travelling 90 \% of the speed of light has a de Broglie wavelength of 2.7 \times 10^{-12} \text{ m}. If the angle of diffraction of the first minima is 5 \degree, what is the nuclear radius of the atom being investigated?

**[2 marks]**

**Question 3: ** Calculate the nuclear radius of gold (atomic number =79).

**[2 marks]**

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