Question

Show that period of an electron in bohr's orbit is proportional to cube of the principal quantum number

Answer 1

Answer:

The time period of the electron in Bohr's orbit of an atom is directly proportional to the cube of the principal quantum number. The centripetal acceleration of electron in Bohr's orbit of an atom is inversely proportional to the fourth power of the principal quantum number.

Answer 2

Complete Question: Show that the time period of the revolution of an electron in Bohr's orbit is proportional to the cube of the principal quantum number (n).

Solution:

Let the mass of an electron be 'm'

Let the value of the radius of the nth orbit be 'r'

Let the velocity of the electron in the nth orbit be 'v'

Let the principal quantum number of the nth orbit be 'n'

Let the atomic number of the atom be 'Z'

The Coulomb's constant is denoted by 'K'

→ According to Bohr's quantization condition, the angular momentum (L) of an electron in an orbit is always an integral multiple of (h)/2π. Here 'h' is a constant known as Planck's constant. The angular momentum (L) of the electron would be equal to n(h)/2π, where n = 1,2,3...and so on. The 'n' is known as Principal quantum number of the orbit.

                                       $\therefore L =n \frac{h}{2\pi}$

                                       $\therefore mvr=n\frac{h}{2\pi} -Eq.(i)$

→ The centripetal force required by an electron to revolve in an orbit will be provided by the electrostatic force of attraction between the nucleus and the electron. Hence the centripetal force (Fc) would be equal to the electrostatic force of attraction (Fe) between the nucleus and the electron.

∵ Number of protons in the nucleus = Z (Atomic number)

∴ Charge on the nucleus = Z(e)

                                        $\therefore F_{c} =F_{e} \\\\\therefore \frac{mv^{2} }{r} =K\frac{(e)(Ze)}{r^{2} }-Eq.(ii)$

→ Squaring the equation (i) and dividing it by the equation (ii):

                                        $\therefore \frac{m^{2}v^{2} r^{2} }{(\frac{mv^{2} }{r} )} =\frac{(n\frac{h}{2\pi} )^{2} }{(K\frac{Ze^{2} }{r^{2} } )}$

                                          $\therefore mr^{3} =\frac{n^{2}h^{2} r^{2} }{4(\pi)^{2}KZe^{2} }$

                                           $\therefore r=\frac{n^{2} h^{2} }{4\pi^{2}mKZe^{2} }$

→ Now putting the value of radius 'r' back in Equation (i) to get the velocity:

                                          $\because mvr=n\frac{h}{2\pi} \\\\\therefore mv(\frac{n^{2} h^{2} }{4\pi^{2}mKZe^{2} })=n\frac{h}{2\pi} \\\\\therefore v=\frac{2\pi KZe^{2} }{nh}$

→ Since the time period (T) in a circular motion is equal to (2πr)/v

                                          $\therefore T=\frac{2\pi r}{v}$

                                          $\therefore T=2\pi(\frac{n^{2} h^{2} }{4\pi^{2} mKZe^{2} } )(\frac{nh }{2\pi KZe^{2} } )$

                                          $\therefore T=n^{3}( \frac{h^{3} }{4\pi^{2}mK^{2} Z^{2}e^{4} } )\\\\\therefore T \propto n^{3}\;(Because\;rest\;all\;are\;constants)$

Hence the time period of the revolution of an electron in Bohr's orbit is proportional to the cube of the Principal quantum number (n).

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