Test Index

CBSE Class 12 Physics 2016 Delhi Set 1 Paper

© examsnet.com
Question : 21 of 26
Marks: +1, -0
(i) State Bohr's quantization condition for defining stationary orbits. How does de Broglie hypothesis explain the stationary orbits?
(ii) Find the relation between the three wavelength λ1,λ2\lambda_1, \lambda_2 and λ3\lambda_3 from the energy level diagram shown below.
Solution:  
(i) Statement of Bohr's quantization condition
de- Broglie explanation of stationary orbits
(ii) Relation between λ1,λ2λ3\lambda_1, \lambda_2 \lambda_3
(i) Only those orbits are stable for which the angular momentum, of revolving electron, is an integral multiple of   h2π\; \frac{h}{2\pi} .
[Alternatively
L=  nh2πL = \; \frac{n h}{2\pi} i.e. angular momentum of orbiting electron is quantized.]
According to de-Broglie hypothesis
Linear momentum (p)=  hλ(p) = \; \frac{h}{\lambda}
And for circular orbit L=rnpL = r_n p where ' rr ' is the radius of quantized orbits.
=  rhλ= \; \frac{r h}{\lambda}Also L=  nh2πL = \; \frac{n h}{2\pi}
∴    rhλ=  nh2π\therefore \;\; \frac{r h}{\lambda} = \; \frac{n h}{2\pi}
⇒2πrn=nλ\Rightarrow 2\pi r_n = n\lambda
∴\therefore Circumference of permitted orbits are integral multiples of the wave-length λ\lambda.
(ii)   EC−EB=  hcλ1\; E_C - E_B = \; \frac{h c}{\lambda_1} .......(i)
  EB−EA=  hcλ2\; E_B - E_A = \; \frac{h c}{\lambda_2} .......(ii)
  EC−EA=  hcλ3\; E_C - E_A = \; \frac{h c}{\lambda_3} .......(iii)
Adding (i) & (ii)
EC−EA=  hcλ1+  hcλ2E_C - E_A = \; \frac{h c}{\lambda_1} + \; \frac{h c}{\lambda_2} ........(iv)
Using equation (iii) and (iv)
  hcλ3  =  hcλ1+  hcλ2\; \frac{h c}{\lambda_3} \; = \; \frac{h c}{\lambda_1} + \; \frac{h c}{\lambda_2}
⇒    1λ3=  1λ1+  1λ2\Rightarrow \;\; \frac{1}{\lambda_3} = \; \frac{1}{\lambda_1} + \; \frac{1}{\lambda_2}
© examsnet.com
Go to Question: