Physics - Modern Physics Question with Solution | TestHub

The fine-structure constant, denoted by $\alpha$, is a fundamental physical constant that characterizes the strength of the electromagnetic interaction. It is a dimensionless quantity. It relates the elementary charge $e$, the speed of light $c$, Planck's constant $h$, and the vacuum permittivity ${ϵ}_0$.

Given the Coulomb constant $K_e=\frac{1}{4\pi {ϵ}_0}$, the electron charge $e$, Planck's reduced constant $\hslash =\frac{h}{2\pi }$, and the speed of light $c$, the fine-structure constant is defined as:

$$\alpha =\frac{K_e{e}^2}{\hslash c}$$

To derive this, we can start by considering the Bohr model of the hydrogen atom. In the Bohr model, the electron orbits the proton at specific energy levels. The speed of the electron in the first Bohr orbit is given by $v=\alpha c$, where $\alpha$ is the fine-structure constant.

The electrostatic force between the electron and the proton is given by Coulomb's law:

$$F=K_e\frac{e^2}{r^2}$$

where $r$ is the radius of the orbit.

The centripetal force required to keep the electron in orbit is:

$$F=\frac{m{v}^2}{r}$$

where $m$ is the mass of the electron.

Equating these two forces, we get:

$$K_e\frac{e^2}{r^2}=\frac{m{v}^2}{r}$$

$$K_e{e}^2=m{v}^{2}r$$

From the Bohr quantization condition, the angular momentum of the electron is quantized:

$$L=mvr=n\hslash$$

For the lowest energy level (n=1), we have:

$$mvr=\hslash$$

$$r=\frac{\hslash }{mv}$$

Substituting this expression for $r$ into the equation $K_e{e}^2=m{v}^{2}r$, we get:

$$K_e{e}^2=m{v}^2\frac{\hslash }{mv}$$

$$K_e{e}^2=v\hslash$$

$$v=\frac{K_e{e}^2}{\hslash }$$

Since $v=\alpha c$, we have:

$$\alpha c=\frac{K_e{e}^2}{\hslash }$$

$$\alpha =\frac{K_e{e}^2}{\hslash c}$$

Therefore, option A is the correct answer: $\frac{K_e{e}^2}{\hslash c}$