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State the postulates of Bohr’s atomic model and derive the expression for the energy of an electron in the atom. - Physics

Answer in Brief

State the postulates of Bohr’s atomic model and derive the expression for the energy of an electron in the atom.

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Solution

The postulates of Bohr's atomic model (for the hydrogen atom):

  1. The electron revolves with a constant speed in a circular orbit around the nucleus. The necessary centripetal force is the Coulomb force of attraction of the positive nuclear charge on the negatively charged electron.
  2. The electron can revolve without radiating energy only in certain orbits, called allowed or stable orbits, in which the angular momentum of the electron is equal to an integral multiple of h/27t, where h is Planck's constant.
  3. Energy is radiated by the electron only when it jumps from one of its orbits to another orbit having lower energy. The energy of the quantum of electromagnetic radiation, i.e., the photon, emitted is equal to the energy difference of the two states.

Consider the electron revolving in the nth orbit around the nucleus of an atom with the atomic number Z. Let m and -e be the mass and the charge of the electron, r the radius of the orbit and v the linear speed of the electron.

According to Bohr's first postulate,

centripetal force on the electron = electrostatic force of attraction exerted on the electron by the nucleus

∴ `"mv"^2/"r" = 1/(4piε_0) = "Ze"^2/"r"^2`   ...(1)

where ε0 is the permittivity of free space.

∴ Kinetic energy (KE) of the electron

`= 1/2 "mv"^2 = "Ze"^2/(8piε_0"r")`   .....(2)

The electric potential due to the nucleus of charge + Ze at a point at a distance r from it is

V = `1/(4piε_0)*"Ze"/"r"`

∴ Potential energy (PE) of the electron

= charge on the electron x electric potential

= `-"e" xx 1/(4piε_0) "Ze"/"r" = - "Ze"^2/(4piε_0"r")`    ....(3)

Hence, the total energy of the electron in the nth orbit is

E = KE + PE = `"-Ze"^2/(4piε_0"r") + "Ze"^2/(8piε_0"r")`

∴ E = `"-Ze"^2/(8piε_0"r")`       ......(4)

This shows that the total energy of the electron in the nth orbit of the atom is inversely proportional to the radius of the orbit as Z, ε0 and e are constants. The radius of the nth orbit of the electron is

r = `(ε_0"h"^2"n"^2)/(pi"mZe"^2)`     ....(5)

where his Planck's constant.

From Eqs. (4) and (5), we get,

`"E"_"n" = - "Ze"^2/(8piε_0)((pi"mZe"^2)/(ε_0"h"^"n"^2)) = - ("mZ"^2"e"^4)/(8ε_0^2"h"^2"n"^2)`     ....(6)

This gives the expression for the energy of the electron in the nth Bohr orbit. The minus sign in the expression shows that the electron is bound to the nucleus by the electrostatic force of attraction.

As m, Z, e, ε0 and h are constant, we get

`"E"_"n" prop 1/"n"^2`

i.e., the energy of the electron in a stationary energy state is discrete and is inversely proportional to the square of the principal quantum number.

Concept: Bohr’s Atomic Model
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APPEARS IN

Balbharati Physics 12th Standard HSC Maharashtra State Board
Chapter 15 Structure of Atoms and Nuclei
Exercises | Q 3 | Page 342
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