Question

In accordance with the Bohr’s model, find the quantum number that characteristics the earth’s revolution around the sun in an orbit of radius $1.5$×${{10}^{11}}$m with the orbital speed 3×${{10}^{4}}$m/s. ( mass of earth=6.0×${{10}^{24}}$kg).

Answer 1

Hint: From the bohr’s model of atom , we know that the formula of angular momentum is mvr=$\dfrac{nh}{2\pi }$, where m is mass, v is the velocity , r is the radius , n is the principal quantum number and h is the Planck’s constant ,thus from this formula, we can easily find the quantum number that characterizes the earth’s revolution.

Complete answer:
First of let’s discuss what is bohr’s model of an atom whose main postulates are as follows:
An atom consists of a small and positively charged nucleus and the negatively charged electrons are revolving around the nucleus in fixed orbits (circular path the nucleus).
The electrons revolve around the nucleus in those fixed orbits and have a definite value of energy and these orbits are also known as energy levels and the energy of the electrons in that particular orbit is fixed and doesn’t change with time. The different energy levels are numbered as 1,2,3,4,5---etc. starting from the nucleus. The energy of each orbit is given by the expression as:
            ${{E}_{n}} =-\dfrac {2{{\pi} ^{2}}m{{e}^ {4}}} {{{n}^ {2}} {{h}^ {2}}} $
Substituting the values, the values of m (mass of the electron), e (charge on the electron) and h (Planck’s constant), we get:
          ${{E}_{n}} =-\dfrac {21.8\times {{10} ^ {-19}}} {{{n}^ {2}}} J/atom$
                =$-\dfrac {1312} {{{n}^ {2}}} kJ/mol$
                =$-\dfrac {13.6} {{{n}^ {2}}} eV/atom$(1eV=1.602 ×${{10} ^ {-19}} $J)
Where n=1,2,3, ----etc. stands for the 1st,2 Nd, 3rd ----etc. levels. The energy level which is closest to the nucleus has the lowest energy and the energy increases as the energy levels increase and so on.
But for hydrogen like particles, the expression for energy is:
          ${{E}_{n}} =-\dfrac {2{{\pi} ^{2}}m{{Z}^ {2}} {{e}^ {4}}} {{{n}^ {2}} {{h}^ {2}}} $
                =$-\dfrac{1312{{Z}^ {2}}} {{{n}^ {2}}} kJ/mol$
Where Z is the atomic number of the element.
The electrons revolve only in those orbits which have the fixed values of energies, hence the electrons in an atom can have only certain definite values of energy and thus, the energy of an electron is quantized. Like the energy, the angular momentum (when a substance rotates about its own axis, it is called as an angular momentum) of an electron also have fixed energy and its value:
              mvr=$\dfrac {Nh} {2\pi} $
where m is the mass of the electron, v is the velocity of the electron, r is the radius of the orbit, h is the Planck’s constant, n is principal quantum number which depicts the energy level to which an electron belongs to i.e. 1,2,3---etc.
 Now, we know that the Radius of the orbit of the Earth around the Sun, r = 1.5 × 1011 m(given).
Orbital speed of the Earth, ν = 3 × 104 m/s(given)
Mass of the Earth, m = 6.0 × 1024 kg(given)
According to the Bohr’s model, discussed above, as we know that, the angular momentum is:
mvr =$\dfrac {Nh} {2\pi} $
Where,
h = Planck’s constant = 6.62 × 10−34 Js
n = Quantum number
∴ n =$\dfrac {mvr2\pi} {h}$

=$\dfrac {2\times 3.14\times 6\times {{10} ^ {24}} \times 3\times {{10} ^ {4}} \times 1.5\times {{10} ^ {11}}} {6.62\times {{10} ^ {-34}}} $
= 25.61x1073
= 2.6 x 1074
Hence, the earth’s revolution around the sun in an orbit of radius $1.5$×${{10}^{11}}$m with the orbital speed 3×${{10}^{4}}$m/s. ( mass of earth=6.0×${{10}^{24}}$kg) is 2.6 × 1074 .

Note:
Don’t get confused in the angular and linear momentum. Angular momentum is the motion of electrons around its own axis whereas linear momentum is the motion of electrons around the nucleus.