Education Journalist | Study Abroad Strategy Lead
Bohr radius was invented by Neils Bohr in 1913. It indicates the distance between the electron and the nucleus in a Hydrogen atom in a ground state. The SI unit of Bohr Radius is 5.29 x 10-11 m.
| Table of Content |
Key Terms: Bohr radius, coulomb’s law, Electron, Atom, Hydrogen, Nucleus, Planck’s constant, angular momentum, velocity, light
Bohr Radius: Explanation
[Click Here for Sample Questions]
Before Bohr we weren’t really sure how electrons managed to stay in the orbit around the nucleus.
Bohr suggested that the atom had quantised energy levels around it that were the only places an electron was allowed to be in.
To derive this he simply postulated that the angular momentum of the electron has to be some integer multiple of Planck’s constant (6.626x10-34 m2kg/s) over 2π.
He then used these assumptions with classical physics to come up with the radius of the hydrogen atom (that turned out to be correct).
Also Read:
| Related Articles | ||
|---|---|---|
| LASER LIGHT | Avogadro's Number | Laser |
| Alpha Particle Mass | Alpha-particle Scattering and Rutherford’s Nuclear Model of Atom | Modern Physics |
Bohr Radius Formula
[Click Here for Sample Questions]
The formula of Bohr radius is
a0 = 4πε0(h/2π)2/mee2
= (h/2π)/mecα
Where,
- ao = Bohr radius.
- me =rest mass of electron.
- εo = permittivity of free space
-
![reduced Planck constant]()
= reduced Planck constant. - c = velocity of light (vacuum).
- α = fine structure constant.
- e = elementary charge.
= reduced Planck constant.
Bohr Radius Derivation: Examples
[Click Here for Sample Questions]
Let’s derive the postulate:
Assume we have some electron around a proton and it’s in a circular orbit undergoing circular motion (hydrogen atom).
Now, the above assumption means there must be a centripetal force acting on the electron and of course we know the centripetal force is coming from the electrical force given by coulomb’s law.
Let’s plug in the values for the charges since one is negative elementary charge and the other is positive elementary charge. (Refer below figure-2)
According to coulomb’s law
F = mv2/r
F = kq1q2/r2 (equation 1)
mv2/ r = ke (-e)/r2
mv2 = k (-e2)/r2
Looking at the magnitude of the force we can take absolute value on both the sides
mv2 = k(e2)/r (equation 2)
Calculating the velocity of the electron
Let’s now come to proving the Bohr’s postulate using De Broglie wavelength.
De Broglie stated that the particles are actually wave-like.
Λ = h/mv (De Broglie wavelength postulate) (equation 3)
Now getting waves in an orbit seems different .Unless, you use standing waves to just fit the waves around the circle (refer below figure 3).
The circumference of the circle would have to be an integer multiple of the wavelength of the electron.
We can substitute in the expression for the De Broglie wavelength and rearrange for the angular momentum: mvr
2πr = nλ
2πr = nλ/mv (equation 4)
mvr = nh/2π (equation 5)
Out of De-Broglie assumption of matter waves comes Bohr’s postulates of quantised angular momentum.
Now, to get the radius of hydrogen atom let’s rearrange for v,
V = nh/2πmr
= n\(h^2\)/ mr (equation 6)
Now plug equation 1 values in equation 6, we get
m (n\(h \)/mr)2 = ke2/r
mn2\(h \)2/m2r2 = ke2/r
n2\(h \)2 /mr = ke2
r = n2\(h \)2/mke2 (equation 7)
Where, r = Bohr radius
m = rest mass of electron
\(h \) = plank constant
The above equation 7 depends on the De-Broglie principle.
Now, plugging any values in equation 7 you will be getting the result around 5.3x1011 m as the radius of the hydrogen atom.
Bohr Radius in Different Units
[Click Here for Sample Questions]
The value of Bohr radius in different units has been tabulated below:
| Units | a0 (Bohr Radius) |
|---|---|
| SI units | 5.29 x 10−11 m |
| Imperial or US units | 2.08 x 10−9 in |
Natural units | 2.68 x 10−4 /eV |
| 3.27 x 1024 \(l\)P |
Uses of Bohr Radius
[Click Here for Sample Questions]
Bohr radius can be used in different units.
In SI unit the Bohr radius is 5.29 x 10-11 m, in US units it is 2.08 x 10-9 in, while in natural units it is 2.68 x 10-4 /eV or 3.27 x 1024l
.
Examples
Real life example of Bohr Radius is while climbing up the ladder you can’t skip a step .You only climb a specific step in space between each ladder. As you move up through the ladder your potential energy will increase gradually, while moving downwards your energy will decrease just like the Bohr radius model.
Things to Remember
- Bohr’s model explains spectra of hydrogenic atoms.
- Neil Bohr’s hypothesis explains the stability of an atom.
- An electron's angular momentum in its orbit is quantized.
- Bohr’s radius is not applied anywhere now because it violates the Heisenberg Uncertainty principle and it can’t also be used in creating diagrams of larger atoms.
Also Read:
| Related Articles | ||
|---|---|---|
| Value of c | Spectral Series | Quantum Mechanics |
| Energy Level | Quantum Physics | Atoms MCQ |
| Atomic Spectra | Atoms Ncert Solutions | Atoms Important Questions |
Sample Questions
Ques. Write the expression for Bohr’s radius in a hydrogen atom. (1 Mark) [Delhi 2010]
Ans. Bohr’s radius in hydrogen atom,
\(r_1 = \frac{\epsilon_0h^2}{\pi me^2} = 0.529 * 10^{-10} m \)
Ques. The value of a ground state energy of a hydrogen atom is -13.6 eV.
(i) Find the energy required to move an electron from the ground state to the first excited state of the atom.
(ii) Determine
(a) the kinetic energy and
(b) orbital radius in the first excited state of the atom. (Given the value of Bohr radius = 0.53 Å) (3 Marks) [Comptt. All India 2014]
Ans. (i) Since, En = E0/n2
∴ E3 = -13.6/(3)2 = -13.6/9 = -1.51 eV
(ii) (a) KE = -(Total energy) = -(-1.51) = 1.51 eV
(b) ? rn = r0 x n2,
∴ r3 = r0 x (3)2 = (0.53) x 9 - 4.77 Å
Ques. In a hydrogen atom, what is the ratio of the radii of the first three Bohr's orbits? (3 marks)
A) 1:4:9
B) 1:2:3
C) 1:2:4
D) 9:4:1
Ans: Atomic number is equal to 1
Hence now the radius of the nth orbit, rn=0.529n2A?
For the first three orbits values are 1, 2 and 3.
As a result, the ratio of the first three circles' radii = r1:r2:r3
=n12:n22:n32
=12:22:32
=1:4:9
Ques. What is the radius of Bohr’s first orbit in He2+? (3marks)
A) 25.64 pm
B) 18.85 pm
C) 26.45 pm
D) 36.85 pm
Ans: Radius of orbit = r
r = n2h2/4π2me2 x1/z
=0.529 x 12/2
=0.2645 Å
In picometers, the radius =0.2645x102=26.45 pm
Ques. The radius of Bohr’s orbit in Li+2 is ______(3 marks)
A) 0.0587 pm
B) 17.63 pm
C) 176.3 pm
D) 0.529 pm
Ans: Radius of orbit
r=n2h2/4π2me2 x 1/z
=0.529 x n2/z
=0.529 x 12/3
=0.1763A?
In picometers, the radius is 0.1763x102=17.63 pm.
Ques. The radius of a hydrogen atom's innermost electron orbit is 5.3 x 10-11 m. What are the radii of the n=3 and n=4 orbit? (6 marks)
Ans. The radius of a hydrogen atom's innermost orbit,
r1=5.3 x 10-11
At n=3, let r2 be the orbit's radius.
It is proportional to the innermost orbit's radius.
r2= (n)2r1
= (3)2x5.3x10-11 m
=4.77x10-10 m
For n=4, we can write the corresponding radius as:
r3= (n)2 r1
= (4)2r1
=16x5.3x10-11 m
=8.48x10-10 m
Hence, the radii of electrons for n=3 and n=4 are 4.77x10-10 m and 8.48x10-10 m respectively.
For Latest Updates on Upcoming Board Exams, Click Here: https://t.me/class_10_12_board_updates
Check-Out:
Comments