Question
The graph between the energy log $E$ of an electron and its de$-$Broglie wavelength log $\lambda$ will be
✓
Answer
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 $-\frac{1}{2}$.
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