The log-log graph between the energy E of an | vedclass

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(C) The de-Broglie wavelength $\lambda$ of an electron with energy $E$ is given by the relation:$\lambda = \frac{h}{\sqrt{2mE}} = \frac{h}{\sqrt{2m}}

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(C) The de-Broglie wavelength $\lambda$ of an electron with energy $E$ is given by the relation:$\lambda = \frac{h}{\sqrt{2mE}} = \frac{h}{\sqrt{2m}} \cdot E^{-1/2}$Taking the logarithm on both sides:$\log \lambda = \log \left( \frac{h}{\sqrt{2m}} \cdot E^{-1/2} \right)$Using the properties of logarithms:$\log \lambda = \log \left( \frac{h}{\sqrt{2m}} \right) + \log (E^{-1/2})$$\log \lambda = -\frac{1}{2} \log E + \log \left( \frac{h}{\sqrt{2m}} \right)$This equation is in the form of a straight line $y = mx + c$,where:$y = \log \lambda$$x = \log E$$m = -1/2$ (slope)$c = \log \left( \frac{h}{\sqrt{2m}} \right)$ (positive intercept on the $y$-axis)Since the slope is negative,the graph is a straight line with a negative slope,which corresponds to the graph shown in option $C$.

The $log-log$ graph between the energy $E$ of an electron and its de-Broglie wavelength $\lambda$ will be

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