A) \[{{10}^{-\,4}}\]
B) \[{{10}^{-6}}\]
C) \[{{10}^{-\,8}}\]
D) \[{{10}^{-10}}\]
Correct Answer: B
Solution :
[b] \[\Delta E.\Delta t\sim h\] \[E=\frac{hc}{\lambda }\text{ }\Delta \Epsilon =\frac{hc}{\lambda }\frac{\Delta \lambda }{\lambda }\] \[\therefore \frac{hc}{\lambda }\frac{\Delta \lambda }{\lambda }.\Delta t\sim h\] Now, \[c=v\lambda \] \[v\Delta \lambda +\lambda \Delta v=0\] \[vD\lambda =-\lambda \Delta v\] \[\therefore \frac{\Delta \lambda }{\lambda }=-\frac{\Delta v}{v}\text{ }\therefore \frac{ch}{\lambda }\frac{\Delta v}{v}\Delta t\sim h\] \[\therefore \frac{\Delta v}{v}\sim \frac{h}{\Delta t}.\frac{\lambda }{hc}\sim \frac{\lambda }{\Delta tc}=\frac{600\times {{10}^{-9}}}{8\times {{10}^{-9}}\times 3\times {{10}^{8}}}\] \[\frac{\Delta v}{v}\sim {{10}^{-6}}\]
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