Question

Calculate the number of waves made by a Bohr’s electron in one complete revolution in its 3rd orbit of H-atom.

Answer 1

Hint: The Bohr model postulates that the electrons surround the positively charged nucleus in specific allowable paths (known as orbits) at fixed energy levels. Orbits exist at higher energy levels from the nucleus. When electrons are returned to a lower energy level, they usually emit energy in the form of light.

Complete step by step solution: According to Bohr’s postulate, electrons revolve only in those orbits in an atom in which the angular momentum is an integral multiple of \[2\Pi h\] where, h = Planck's constant. According to this postulate, angular momentum of nth orbit = $\dfrac{{nh}}{{2\Pi }}$
$mvr = \dfrac{{nh}}{{2\Pi }}$ (m = mass of the particle, v = velocity, r = radius)
de Broglie postulated that the particles can exhibit the properties of waves.
According to de Broglie:
$mv = \dfrac{h}{\lambda }$ (λ = wavelength)
Substituting the value of mv in the previous equation:
$
  \dfrac{h}{\lambda }r = \dfrac{{nh}}{{2\Pi }} \\
  \therefore n\lambda = 2\Pi r \\
$
It is given that n = 3, thus:
$
  3\lambda = 2\Pi r \\
 \Rightarrow \lambda = \dfrac{{2\Pi r}}{3} \\
$
We know that:
$Number{\text{ }}of{\text{ }}waves{\text{ }} = {\text{ }}\dfrac{{Circumference{\text{ }}of{\text{ }}electron{\text{ }}orbit}}{{Wavelength}} = \dfrac{{2\Pi r}}{\lambda }$
Now, substitute the value of λ in this equation to calculate the number of waves:
$Number{\text{ }}of{\text{ }}waves = \dfrac{{2\Pi r}}{{\dfrac{{2\Pi r}}{3}}} = 3$

Hence, the number of waves made by a Bohr’s electron in one complete revolution in its 3rd orbit of H-atom is 3.

Note: You can directly know the answer of this question without doing this lengthy calculation. Always remember that the number of waves is the same as the principal quantum number i.e. n. For example, if n = 3, number of waves = 3 and if n = 6, number of waves = 6.