c
As we know that, wavelength of a particle

\(\lambda=\frac{h}{\sqrt{2 m E}}=\frac{h}{\sqrt{2 m}} \cdot \frac{1}{\sqrt{E}}\)

\(\Rightarrow \log \lambda=\log \left(\frac{h}{\sqrt{2 m}} \cdot \frac{1}{E}\right)\)

\(\Rightarrow \log \lambda=\log \frac{h}{\sqrt{2 m}}+\log \frac{1}{E^{1 / 2}}\)

\(\Rightarrow \log \lambda=\log \left(\frac{h}{\sqrt{2 m}}\right)-\frac{1}{2} \log E\)

\(\Rightarrow \log \lambda=-\frac{1}{2} \log E+\log \frac{h}{\sqrt{2 m}}\)

So, equation of straight line is \(y=m x+c\).

This is equation of line with slope \(-\frac{1}{2}\).