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

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(C) As we know that,the de$-$Broglie wavelength of a particle is given by:$\lambda = \frac{h}{\sqrt{2 m E}} = \frac{h}{\sqrt{2 m}} \cdot E^{-1/2}$Taki

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(C) As we know that,the de$-$Broglie wavelength of a particle is given by:$\lambda = \frac{h}{\sqrt{2 m E}} = \frac{h}{\sqrt{2 m}} \cdot E^{-1/2}$Taking the logarithm on both sides:$\log \lambda = \log \left( \frac{h}{\sqrt{2 m}} \cdot E^{-1/2} \right)$Using the property $\log(ab) = \log a + \log b$ and $\log(a^n) = n \log a$:$\log \lambda = \log \left( \frac{h}{\sqrt{2 m}} \right) + \log(E^{-1/2})$$\log \lambda = \log \left( \frac{h}{\sqrt{2 m}} \right) - \frac{1}{2} \log E$Rearranging this into the form of a straight line equation $y = mx + c$:$\log \lambda = -\frac{1}{2} \log E + \log \left( \frac{h}{\sqrt{2 m}} \right)$Here,the slope $m = -1/2$,which is negative. This represents a straight line with a negative slope and a positive intercept on the $\log \lambda$ axis. Therefore,the correct graph is shown in option $C$.

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

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