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: (A)

As, $\lambda =\frac{h}{\sqrt{2mE}}=\frac{h}{\sqrt{2m}}\frac{1}{\sqrt{E}}$
Taking log on both the sides
$\log \lambda =\log \frac{h}{\sqrt{2m}}+\log \frac{1}{\sqrt{E}}$
$(\because \log (AB)=\log A+\log B)$
$\log \lambda =\log \frac{h}{\sqrt{2m}}-\frac{1}{2}\log E$
or $\log \lambda =-\frac{1}{2}\log E+\log \frac{h}{\sqrt{2m}}$
The equation of straight line having slope $\left(-\frac{1}{2}\right)$ and positive intercept on log $\lambda$ axis