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Atoms — Physics STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 SciencePhysicsAtoms3 Marks
Question
  1. State Bohr’s quantization condition for defining stationary orbits. How does de Broglie hypothesis explain the stationary orbits?
  2. Find the relation between the three wavelengths$\lambda_{1} , \lambda_{2}$ and$\lambda_{3}$ from the energy level diagram shown below.
✓

Answer

Only those orbits are stable for which the angular momentum, of revolving electron, is an integral multiple of $\frac{\text{h}}{2\pi}.$
Alternate Answer
$[\text{L} = \frac{\text{nh}}{2\pi}$ i.e. angular momentum of orbiting electron is quantised.]
According to de Broglie hypothesis
Linear momentum $(\text{P}) = \frac{\text{h}}{\lambda}$
And for circular orbit $\text{L} =\text{r}_{n}\text{p}$ where $' \text{r}_{n} '$ is the radius of quantized orbits.
$ = \frac{\text{rh}}{\lambda}$
Also $\text{L} = \frac{\text{nh}}{2\pi}$
$\therefore\frac{\text{rh}}{\lambda} = \frac{\text{nh}}{2\pi}$
$\Rightarrow2\pi\text{r}_{n} = \text{n}\lambda$
$\therefore$ Circumference of permitted orbits are integral multiples of the wavelength $\lambda$
$\text{E}_{c} - \text{E}_{B} = \frac{\text{hc}}{\lambda}_{1}$ . . . . (i)
$\text{E}_{B} - \text{E}_{A} =\frac{\text{hc}}{\lambda}_{2}$ . . . . (ii)
$\text{E}_{c} - \text{E}_{A} = \frac{\text{hc}}{\lambda}_{a}$ . . . . . (iii)
Adding (i) & (ii)
$\text{E}_{c} - \text{E}_{A} = \frac{\text{hc}}{\lambda}_{1} + \frac{\text{hc}}{\lambda}_{2}$ . . . . . (vi)
Using equation (iii) and (iv)
$\frac{\text{hc}}{\lambda}_{3} = \frac{\text{hc}}{\lambda}_{1} + \frac{\text{hc}}{\lambda}_{2} = > \frac{1}{\lambda}_{3} = \frac{1}{\lambda}_{1} + \frac{1}{\lambda}_{2}$

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