Bohr Radius

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What is Bohr Radius? 

Bohr radius is a physical constant in atomic physics. It explains the most probable distance between the nucleus and the electron at the ground state of a hydrogen atom.

The symbol to denote the Bohr radius is ao or rBohr. This constant is named after the scientist Niels Bohr as he formed the Bohr model.

Bohr put forward the Bohr model of atomic structure in 1913. He stated that electrons revolve around the central nucleus under electrostatic attraction. The derivation states that electrons possess the orbital angular momentum in multiple forms of integers of the reduced Planck constant.

Bohr chose Hydrogen as it is the simplest atom. He discovered that a single electron rotates around the nucleus.

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Bohr Radius Explained

Danish Physicist and Philosopher Niels Bohr put forward the Bohr model of the atom. He proposed that atoms are the composition of small and dense nuclei with positively and negatively charged electrons. 

The electrons are orbiting around the nuclei in circular paths. From current observations, this theory is said to be an oversimplified one. He also stated that the electrons are surrounded and roam around the nucleus in an orbicular probability zone- like shells.

However, the importance of the Bohr radius is quite high. It states that the smallest average radius is usually attainable by a neutral atom.

We can calculate the Bohr radius of reduced mass in the hydrogen atom. The equation is given as:

a\[_{0}^{\ast }\] = \[\frac{\lambda_{p}+\lambda_{e}}{2\pi \alpha}\] = \[\frac{m_{e}}{\mu}\]a\[_{0}\] = \[\frac{h}{\mu c\alpha }\] = \[\frac{4\pi \epsilon_{0}h^{2}}{\mu\mid q_{e}\parallel q_{p}\mid }\]

Here, 

\[\lambda\]\[_{p}\] = Compton wavelength of the proton,

\[\lambda\]\[_{e}\] = Compton wavelength of the electron,

h = reduced Planck's constant,

\[\alpha\] = fine structure constant,

c = speed of light,

\[\mu\] = reduced mass of the electron/proton system,

\[\epsilon\] = permittivity of free space,

\[\mid\] q\[_{e}\] \[\mid\] = the magnitude of the charge of the electron,

\[\mid\] q\[_{p}\] \[\mid\] = proton’s magnitude

From the first expression, the effect of the reduced mass is calculated by using the incremented Compton wavelength. To calculate Compton wavelength, one must add the Compton wavelength of electron and proton.

The application of reduced mass is characteristically a traditional overview of the two-body problem. When we are outside of the calculation, the mass of the circling body is smaller than the mass of the revolving body.

It is observed that the electron mass is bigger (slightly) than the reduced mass of the electron or proton system.


How to Find Bohr Radius?

Here is the explanation to find the Radius of Nth Bohr Orbit using Bohr’s atomic model:

We know that the centripetal force = \[\frac{mv^{2}}{r}\] 

Electrostatic force = \[\frac{1}{4\pi \epsilon_{0}}\] . \[\frac{z.e^{2}}{r^{2}}\]

In a Hydrogen atom, electrostatic attraction force = centrifugal force.  

So, \[\frac{1}{4\pi \epsilon_{0}}\] . \[\frac{z.e^{2}}{r^{2}}\] = \[\frac{mv^{2}}{r}\]

v\[^{2}\] = \[\frac{1}{4\pi \epsilon_{0}}\] . \[\frac{ze^{2}}{mr}\].....[1]

From, Bohr’s quantum equation,

L = mvr = nħ…..[2]

Where ħ = \[\frac{h}{2 \pi }\]

By solving equation 2, we get:

v = \[\frac{nħ}{mr}\]

Putting the value of v in eqn 1, we obtain:

(\[\frac{nħ}{mr}\])\[^{2}\] = \[\frac{1}{4\pi \epsilon_{0}}\] . \[\frac{ze^{2}}{mr}\]

⇒ \[\frac{n^{2}ħ^{2}}{m^{2}r^{2}}\] = \[\frac{1}{4\pi \epsilon_{0}}\] . \[\frac{ze^{2}}{mr}\]

Radius, r = \[\frac{4\pi \epsilon_{0}n^{2}ħ^{2}}{mze^{2}}\]...(3)

We know the value of 4\[\pi\]\[\epsilon\]\[_{0}\] is : 1/9 х 10\[^{9}\]

Reduced Plank’s constant has a value which is: ħ = \[\frac{h}{2 \pi}\] = \[\frac{6.625 \times 10^{-34}}{2 \pi}\]

To calculate the radius for the Hydrogen atom

n = 1 

mass of electron m = 9.11 * 10-31Kg

z = 1

e = 1.6 * 10-19C

If we put all the values in equation 3, we get:

The radius of Hydrogen Atom in Meters, r = 0.529 m

The Bohr orbit radius for the hydrogen atom is approximately = 5.2917721067 * 10-11 m


Bohr Radius in Different Units

The table stated below has the value of the Bohr radius for different units:


Units

(a0) Bohr radius

SI units

5.29×10−11 m

Imperial or US units

2.08×10−9 in

Natural units

2.68×10−4 /eV

3.27×1024 ℓP


Bohr Radius Formula

Here is the formula for calculating the Bohr radius:

r = \[\frac{4 \pi \epsilon _{0} . n^{2}ħ^{2}}{mze^{2}}\]

Here, 

ao = Bohr radius.

me = rest mass of electron.

c = velocity of light in vacuum.

εo = permittivity of the free space.

α = fine structure constant.

e = elementary charge.

ħ = reduced Planck constant = h/2π.

To express the radius of Bohr orbit in Gaussian units, we can write it as:

a\[_{0}\] = \[\frac{(\frac{h}{2 \pi })^{2}}{m_{e}e^{2}}\]


Some Important Uses of Bohr’s Radius

Nowadays, the Bohr model is not very functional in Physics. However, it is highly used for its promising occurrence in manipulating other important physical constants, such as: 

  • The fine structure constant

  • Atomic unit

Schrodinger equation superseded the Bohr model of the atom using an electron probability cloud. This is a complicated process. The complication became more when spin and quantum vacuum effects were initiated to harvest fine structure and hyperfine structure.

FAQ (Frequently Asked Questions)

Q1. What Do You Learn from the Bohr Model?

Ans: Bohr model taught us how electrons travel in distinct circular orbits around the nucleus. The orbits are named with the letter ‘n’, and the value of n is an integer. The transfer of electrons is possible by emitting and absorbing energy.

Q2. What Do You Mean By Electrostatic Attraction?

Ans: The electrostatic attraction is an attracting or repelling force that refers to long-range interaction occurring between the atoms of different particles being differently charged or uncharged. The electrostatic attraction is a stronger attraction among the atoms of the particle.

Q3. Write a Few Lines about Static Electricity.

Ans: Static electricity is found within or on the superficial of a material with an imbalance distribution of charge. It remains in the surface position until there is no disturbance from an outside electric current or electric discharge.

Q4. Calculate the Radius of the Second Bohr Orbit of Lithium-ion (Li+2) in Terms of Bohr Radius aₒ.

Ans: We know that for Lithium-ion (Li+2), the values of n and z are:

n = 2, z = 3

The formula of Bohr radius, r = a₀.n²/z 

=  a₀.(2)²/3 = 4a₀/3