Question

Find out the number of waves made by a Bohr electron in one complete revolution in its third orbit.
A. $3$
B. $6$
C. $9$
D. $1$

Answer 1

Hint: Bohr suggested the model of an atom and its postulates surpass the disadvantages of Rutherford’s model and describe the line spectrum of hydrogen. Bohr's model of an atom mostly says about the energy levels in the atom. The postulates of the Bohr model suggest that the electrons encircle the positively charged nucleus in definite allowable tracks called orbits at fixed energy levels. Orbits occur at higher energy levels from the nucleus. When electrons come back to a lower energy level, they typically discharge energy in the form of light.

Complete step by step solution:
According to the postulate of Bohr’s model, electrons circle merely in those orbits in an atom in which the angular momentum is an integral multiple of $2\pi h$ where\[{\text{ }}h{\text{ }} = \] Planck's constant. According to this, the angular momentum of the $n$ th orbit $ = \dfrac{{nh}}{{2\pi }}$
$mvr = \dfrac{{nh}}{{2\pi }}$
where \[m = \] mass of the particle,
\[v = \] velocity,
\[r = \] radius
de Broglie suggested that the particles can show the properties of waves.
According to de Broglie:
$mv = \dfrac{h}{\lambda }$ (\[\lambda = \]wavelength)
Replacing the value of \[mv\]in the previous equation
  $\dfrac{h}{\lambda }r = \dfrac{{nh}}{{2\pi }}{\text{ }}$
$n\lambda = 2\pi r{\text{ }}$
It is given that\[n{\text{ }} = {\text{ }}3\], thus
  $3\lambda = 2\pi r{\text{ }}$
$ \Rightarrow \lambda = \dfrac{{2\pi r}}{3}{\text{ }}$
We know that:
${\text{Number of waves = }}\dfrac{{{\text{Circumference of electron orbit}}}}{{{\text{Wavelength}}}}{\text{ = }}\dfrac{{2\pi r}}{\lambda }$
Now, replace the value of \[\lambda \] in this equation to estimate the number of waves:
${\text{Number of waves}} = \dfrac{{2\pi r}}{{\dfrac{{2\pi r}}{3}}} = 3$
Therefore, the number of waves produced by a Bohr’s electron in one complete revolution in its \[3rd\] orbit of H-atom is $3$ .
Hence option A is correct.

Note:
We can also find the solution directly without doing any lengthy calculations. At all times remember that the number of waves is equal to the principal quantum number i.e. n. For example, if \[n = 3\] , then the number of waves = $3$ and if $n = 6$ , then the number of waves = $6$ .