Question

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

• A.
• B.
• C.
• D.

Answer 1

Answer: (C)

$\lambda =\frac{h}{\sqrt{2mE}}=\frac{h}{\sqrt{2m}}.\frac{1}{\sqrt{E}}$
$\log \lambda =\log \frac{h}{\sqrt{2m}}+\log \frac{1}{\sqrt{E}}$
$\Rightarrow \log \lambda =\log \frac{h}{\sqrt{2m}}-\frac{1}{2}\log E$
$\Rightarrow \log \lambda =-\frac{1}{2}\log E+\log \frac{h}{\sqrt{2m}}$
This is the equation of straight line having slope $(-1/2)$ and positive intercept on $\log \lambda$ axis.