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State the postulates of Bohr’s atomic model. Hence show the energy of electrons varies inversely to the square of the principal quantum number.

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Question

State the postulates of Bohr’s atomic model. Hence show the energy of electrons varies inversely to the square of the principal quantum number. 

Answer in Brief
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Solution

Bohr’s three postulates are: 

  1. In a hydrogen atom, the electron revolves around the nucleus in a fixed circular orbit with constant speed.
  2. The radius of the orbit of an electron can only take certain fixed values such that the angular momentum of the electron in these orbits is an integral multiple of `"h"/(2π)`, h being the Planck’s constant.
  3. An electron can make a transition from one of its orbits to another orbit having lower energy. In doing so, it emits a photon of energy equal to the difference in its energies in the two orbits.

Expression for the energy of an electron in the nth orbit of Bohr’s hydrogen atom:  

  1. Kinetic energy:
    Let, me = mass of the electron  
    rn = radius of nth orbit of Bohr’s hydrogen atom  
    vn = velocity of electron
    −e = charge of the electron
    +e = charge on the nucleus
    Z = a number of electrons in an atom.
    According to Bohr’s first postulate,  
    `("m"_"e""V"_"n"^2)/"r"_"n" = 1/(4piepsilon_0) xx ("Ze"^2)/("r"_"n"^2)`
    where, `epsilon_0` is permittivity of free space.
    ∴ `"m"_"e""v"_"n"^2 = "Z"/(4piepsilon_0) xx "e"^2/"r"_"n"` ….(1)
    The revolving electron in the circular orbit has linear speed and hence it possesses kinetic energy.
    It is given by, K.E = `1/2 "m"_"e""v"_"n"^2`
    ∴ K.E = `1/2 xx ("Z"/(4piepsilon_0) xx "e"^2/"r"_"n")`  ….[From equation (1)]
    ∴ K.E = `"Ze"^2/(8piepsilon_0"r"_"n")` .…(2)
  2. Potential energy:
    The potential energy of the electron is given by, P.E = V(−e)
    where,
    V = electric potential at any point due to charge on the nucleus
    − e = charge on the electron.
    In this case,
    ∴ P.E = `1/(4piepsilon_0) xx "e"/"r"_"n" xx (-"Ze")`
    ∴ P.E = `1/(4piepsilon_0) xx (-"Ze"^2)/"r"_"n"`
    ∴ P.E = −`("Ze"^2)/(4piepsilon_0"r"_"n")` ….(3)
  3. Total energy:
    The total energy of the electron in any orbit is its sum of P.E and K.E.
    ∴ T.E = K.E + P.E
    = `("Ze"^2/(8piepsilon_0"r"_"n")) + (-"Ze"^2/(4piepsilon_0"r"_"n"))` ….[From equations (2) and (3)]
    ∴ T.E = `-"Ze"^2/(8piepsilon_0"r"_"n")`  ….(4)
  4. But, rn = `((epsilon_0"h"^2)/(pi"m"_"e""Ze"^2)) xx "n"^2`
    Substituting for rn in equation (4), 
    ∴ T.E = `−1/(8piepsilon_0) xx "Ze"^2/(((epsilon_0"h"^2)/(pi"m"_"e""Ze"^2))"n"^2)`
    = `-1/(8piepsilon_0) xx ("Z"^2"e"^2pi"m"_"e""e"^2)/(epsilon_0"h"^2"n"^2)`
    ∴ T.E = −`("m"_"e""Z"^2"e"^4)/(8epsilon_0^2"h"^2) xx 1/"n"^2` ….(5)
    ⇒ T.E. ∝ `1/"n"^2` 
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Chapter 15: Structure of Atoms and Nuclei - Long Answer

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SCERT Maharashtra Physics [English] 12 Standard HSC
Chapter 15 Structure of Atoms and Nuclei
Long Answer | Q 1
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