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