Question

(a) State Bohr's postulate to define stable orbits in hydrogen atom. How does de Broglie's hypothesis explain the stability of these orbits ?
(b) A hydrogen atom initially in the ground state absorbs a photon which excites it to the n = 4 level. Estimate the frequency of the photon.

Answer 1

(a) Bohr's postulate for stable orbits in hydrogen atom : An electron can revolve only in those circular orbits in which its angular momentum is an integral multiple of $\frac{h}{2\pi }$.
If n is the principal quantum number of orbit, then an electron can revolve only in certain orbits or definite radii. These are called stable orbits.

De Broglie explanation of stability of orbits:
According to de Broglie, orbiting electron around the nucleus is associated with a stationary wave. Electron wave is a circular standing wave. Since destructive interference will occur if a standing wave does not close upon itself, only those de Broglie waves exist for which the circumference of circular orbit contains a whole number of wavelengths i.e., for orbit circumference of $n^{th}$ orbit as $2\pi r_n$,

$2\pi r_n=n\lambda$

$2\pi r_n=n\left(\frac{h}{mv}\right)$

or $mv{r}_n=n\left(\frac{h}{2\pi }\right)$

which is the quantum condition proposed by Bohr.

(b) In ground state, n =1
In excited state, n =4
We know,

$\frac{1}{\lambda }=R_H[\frac{1}{(1{)}^2}-\frac{1}{(4{)}^2}]$

$\frac{1}{\lambda }=1.097\times {10}^7\times \frac{15}{16}=10284375$

= $1.028\times {10}^7{m}^{-1}$

Frequency, v =$\frac{c}{\lambda }=3\times {10}^8\times 1.028\times {10}^7$

= $3.09\times {10}^{15}Hz$