Question

plz answer this question with valid explanation

Answer 1

The answer is Option (a).


This question involves the concepts of Heisenberg's Uncetainty Principle and the De-Broglie Hypothesis.


The De-Broglie Hypothesis states that all matter particles also behave like waves. The De-Broglie wavelength of matter particle is given by:


$\lambda = \frac{h}{p}$
where p is momentum. 


Also, Heisenberg's Uncertainty Principle gives us that there is an inherent limit of precision with which things can be measured. 

The product of uncertainties in position and momentum is never less than a certain value.


The mathematical form is:

$\Delta x . \Delta p \geqslant \frac{h}{4\pi}$


Looking at the options, we realize that momentum p is not there. But $\lambda$ is present. So, we must somehow find a relation between momentum and De-Broglie Wavelength. 

We again see the De-Broglie Equation:

$\lambda = \frac{h}{p} \\ \\ \\ \implies p = \frac{h}{\lambda} \\ \\ \\ \implies p = h \times \lambda^{-1} \\ \\ \\ \text{Differentiating with respect to } \lambda \\ \\ \\ \implies \frac{dp}{d\lambda} = h \times (-\lambda^{-1-1}) \\ \\ \\ \implies \frac{dp}{d\lambda} = -\frac{h}{\lambda^2}$

Now, $dp$ and $d\lambda$ refer to very small changes in the quantities. 

They can be called the uncertainties in those quantities as well. So we can write:

$\frac{dp}{d\lambda} = \frac{\text{Change in p}}{\text{Change in }\lambda} \\ \\ \\ \implies \frac{dp}{d\lambda} = \frac{\Delta p}{\Delta \lambda} \\ \\ \\ \implies \frac{\Delta p}{\Delta \lambda} = -\frac{h}{\lambda^2} \\ \\ \\ \implies \Delta p = -\frac{h}{\lambda^2} \, \Delta \lambda$

So, magnitude of uncertainty in momentum will be:

$\Delta p = \frac{h}{\lambda^2} \, \Delta \lambda$



Now, we use this in the Heisenberg's Uncertainty Relation:

$\Delta x . \Delta p \geqslant \frac{h}{4\pi} \\ \\ \\ \implies \Delta x . \left( \frac{h}{\lambda^2} \right) \Delta \lambda \geqslant \frac{h}{4\pi} \\ \\ \\ \implies \frac{\Delta \lambda . \Delta x}{\lambda^2} \geqslant \frac{1}{4\pi} \\ \\ \\ \implies \boxed{\Delta \lambda . \Delta x \geqslant \frac{\lambda^2}{4\pi}}$


Thus, clearly, the answer is Option (a).