Many modern technologies, such as computers and smartphones, are built on the principles of Physics Topics such as quantum mechanics and information theory.
What is the Radius of Bohr orbit?
According to this condition, the radii of different Bohr orbits and orbital speeds of electrons are different. In case of n-th orbit,
mvnrn = \(\frac{n h}{2 \pi}\); where n = 1, 2, 3,…….
According to this condition,
\(v_n^2\) = \(\frac{n^2 h^2}{4 \pi^2 m^2 r_n^2}\) ….. (1)
Radius of the n-th Bohr orbit (rn): Electrostatic force of attraction between electron and nucleus of n th orbit = \(\frac{1}{4 \pi \epsilon_0} \cdot \frac{Z e^2}{r_n^2}\)
This force of attraction supplies the necessary centripetal force \(\frac{m v_n^2}{r_n}\) to the electron so that it can revolve in a circular orbit.
So, \(\frac{m v_n^2}{r_n}\) = \(\frac{1}{4 \pi \epsilon_0} \cdot \frac{Z e^2}{r_n^2}\)
or, \(v_n^2\) = \(\) ….. (2)
or, rn = \(\frac{\epsilon_0 n^2 h^2}{\pi m Z e^2}\) …. (3)
This is the expression of the radius of n-th Bohr orbit.
Total energy of the electron in n-th orbit (En): Kinetic energy of the electron in n-th orbit,
Ek = \(\frac{1}{2} m v_n^2\) = \(\frac{1}{2} \cdot \frac{1}{4 \pi \epsilon_0} \cdot \frac{Z e^2}{r_n}\) = \(\frac{1}{4 \pi \epsilon_0} \cdot \frac{Z e^2}{2 r_n}\)
[putting the value of \(v_n^2\) from equation (2)]
Now, we know that the potential energy of an electron in n-th orbit due to electrostatic force of attraction,
Ep = \(-\frac{1}{4 \pi \epsilon_0}, \frac{Z e^2}{r_n}\)
So, total energy of the electron in n-th orbit,
En = Ek + Ep = \(\frac{1}{4 \pi \epsilon_0} \cdot \frac{Z e^2}{r_n}\)(\(\frac{1}{2}\) – 1) = –\(\frac{1}{4 \pi \epsilon_0} \cdot \frac{Z e^2}{2 r_n}\) ….. (4)
Hence, Ek = -En, and Ep = 2En
Substituting the value of rn, from equation (3) in equation (4), we get,
En = –\(\frac{1}{4 \pi \epsilon_0} \cdot \frac{Z e^2 \cdot \pi m Z e^2}{2 \cdot \epsilon_0 n^2 h^2}\) = –\(\frac{m e^4 Z^2}{8 \epsilon_0^2 n^2 h^2}\) ….. (5)
Hydrogen atom: The structure of hydrogen atom is the simplest of all. For hydrogen Z = 1 , i.e., the electric charge of its nucleus is +e and only one electron having charge -e revolves around it. All the information obtained from Bohr’s theory for this atom are given below.
Radius of n-th Bohr orbit: Substituting Z = 1 in the equation (3) we get,
rn = \(\frac{\epsilon_0 n^2 h^2}{\pi m e^2}\) ….. (6)
Here, ϵ0 = permittivity of vacuum
= 8.854 × 10-12C2 ᐧ N-1 ᐧ m-2;
h = Planck’s constant = 6.63 × 10-34 J ᐧ s;
m = mass of electron = 9.1 × 10-31 kg,
e = charge of an electron = 1.6 × 10-19 C
Hence, rn ∝ n2
The information obtained about Bohr orbits from equation (6) is:
i) For n = 1, r becomes minimum, i.e., the orbit lies closest to the nucleus. This is known as the first Bohr orbit or K – shell of the atom. The radius of this orbit is called the first Bohr radius and is denoted by the sign a0, i.e., r1 = a0.
Putting n = 1 in equation (6), we get,
a0 = \(\frac{\epsilon_0 h^2}{\pi m e^2}\) …. (7)
Substituting the values of the constants in this expression we get,
a0 = 0.53 × 10-10 m = 0.53 Å = 0.053 nm
Angstrom and nanometre units:
1 angstrom (Å) = 10-10m; 1 nanometre (nm) = 10-9m
So, 1 nm = 10 Å or, 1Å = 0.1 nm. Nowadays nm unit is more commonly used instead of Å in SI.
ii) Using the value of a0 in equation (6), we get,
rn = n2a0 ……. (8)
Substituting n = 2, 3, 4, in this equation, the radii of the orbits, L, M, N, … are obtained, respectively.
Since, rn ∝ n2; the ratio of the radii of the orbits K, L, M, N, …. is 1 : 4 : 9 : 16….. .
For example,
radius of the L – orbit, r2 = 22 ᐧ a0 = 4 × 0.53 = 2.12Å ;
radius of the M – orbit, r3 = 32 ᐧ a0 = 9 × 0.53 = 4.77Å; etc.
iii) The orbits having radii a0, 4a0, 9a0, ……. are permissible and no orbit exists in between them. For example, radii of the first and the second Bohr orbits are 0.53Å and 2.12Å, respectively. So, no electron can revolve along any circular path having radius > 0.53 Å but < 2.12 Å. So, the Bohr
orbits are discrete, due to integral values of n. Thus n is called the principal quantum number.
Energy of the electron in n -th orbit (En): Substituting the value of rn, from equation (8) in equation (4), we get (for hydrogen, Z = 1),
En = –\(\frac{1}{4 \pi \epsilon_0} \cdot \frac{e^2}{2 n^2 a_0}\) …. (9)
This is the expression for total energy of the electron revolving in the n-th orbit of a hydrogen atom. The inferences made from equation (9) are:
i) The value of En is negative; hence higher the value of n, i.e., greater the distance greater the energy of the electron. So in the first Bohr orbit or in the K-orbit, the energy of electron is the least. This is called the ground state of hydrogen atom. Putting n = 1 in equation (9), we get,
E1 = –\(\frac{1}{4 \pi \epsilon_0} \cdot \frac{e^2}{2 a_0}\) …… (10)
Using the values of the constants we get, E1 = -13.6eV; hence, the ground state energy of hydrogen atom = -13.6eV. Again, putting the value of a0 from equation (7) into equation (10), we get,
E1 = –\(\frac{1}{4 \pi \epsilon_0} \cdot \frac{e^2 \pi m e^2}{2 \epsilon_0 h^2}\) = –\(\frac{m e^4}{8 \epsilon_0^2 h^2}\)
= –\(\frac{m e^4}{8 \epsilon_0^2 c h^3}\) ᐧ ch = -Rch ……. (11)
[where c is the speed of light in vacuum]
where, R = \(\frac{m e^4}{8 \epsilon_0^2 c h^3}\) = constant ……. (12)
Here, R is a constant called Rydberg constant. Substituting the accepted values of the physical constants, the value of R comes out to be 1.09737 × 107m-1.
ii) From equations (9), (10) and (11), we get,
En = –\(\frac{R c h}{n^2}\) = –\(\frac{13.6 \mathrm{eV}}{n^2}\) …. (13)
or, En ∝ \(\frac{1}{n^2}\)
Thus the energies in the orbits K, L, M, N, …….. are as 1 : \(\frac{1}{4}\) : \(\frac{1}{9}\) : \(\frac{1}{16}]\) : ……… . So, in those orbits of hydrogen atom, the energies are E1 = -13.6eV, E2 = -3.4eV, E3 = -1.51eV, E4 = -0.85eV, ………, respectively. Hence, these energy levels are discrete, i.e., the electron cannot exist in any intermediate energy state.
Clearly, as n increases, E becomes less negative i.e., the energy increases. The energy levels of hydrogen atom are represented in the energy level diagram [Fig. 1.6]. The highest energy state corresponds to n = ∞ and has energy,
E∞ = –\(\frac{13.6}{\infty^2}\) = 0 eV
Note that an electron can have any total energy above E∞ = 0 eV. In such a situation, the electron is free and there is a continuum of energy state above E∞ = 0 eV
Speed of the electron in n-th orbit (vn): In case of, hydrogen atom, putting Z = 1 , we get from equation (2),
\(v_n^2\) ∝ \(\frac{1}{r_n}\) or, vn ∝ \(\frac{1}{\sqrt{r_n}}\)
Again from equation (3), we get
rn ∝ n2 so, vn ∝ \(\frac{1}{n}\)
Hence, ratio of the speeds of electron in K, L, M, N ….. orbit is,
v1 : v2 : v3 : ….. = 1 : \(\frac{1}{2}\) : \(\frac{1}{3}\) : ………
Hence speed of electron is heighest in first Bohr orbit.
In case of hydrogen item, putting Z = 1 in equation (2), we get,
α is called Sommerfeld’s fine structure constant : Substituting the accepted values of the physical constants, the value of α comes out to be α = \(\frac{1}{137}\) approximately.
∴ vn = \(\frac{c}{137 n}\)
For the first Bohr orbit n = 1, the speed of an electron,
v1 = \(\frac{c}{137}\) = 2.817 × 106m ᐧ s-1
This speed is approximately \(\frac{1}{137}\) part of the speed of light (c) in vacuum.
So, the speed of electron in the next orbits are-
v2 = \(\frac{1}{2}\)v1 = \(\frac{c}{274}\) ; v3 = \(\frac{1}{3}\)v1 = \(\frac{c}{411}\) ; etc
Orbital angular momentum of the electron in n-th orbit (Ln) : From Bohr’s quantum condition we directly obtain,
Ln = \(\frac{n h}{2 \pi}\) ; where n = 1, 2, 3, …..
Naturally, Ln ∝ n, i.e., L1 : L2 : L3 : …….. = 1 : 2 : 3 : ……..
So, greater the distance of the orbits, greater will be the value of Ln.
For example,
L1 = \(\frac{h}{2 \pi}\) = \(\frac{6.63 \times 10^{-34}}{2 \pi}\) = 1.06 × 10-34 J ᐧ s ;
L2 = 2L1 = 2.12 × 10-34 J ᐧ s ; etc.