MCQ
The $log-log$ graph between the energy $E$ of an electron and its de-Broglie wavelength $\lambda $ will be
- A
- B
- ✓
- D
✓
Answer
Correct option: C.
c
(c) $\lambda = \frac{h}{{\sqrt {2mE} }} = \frac{h}{{\sqrt {2m} }} \cdot \frac{1}{{\sqrt E }}$.
(c) $\lambda = \frac{h}{{\sqrt {2mE} }} = \frac{h}{{\sqrt {2m} }} \cdot \frac{1}{{\sqrt E }}$.
Taking $log$ of both sides $\log \lambda = \log \frac{h}{{\sqrt 2 m}} + \log \frac{1}{{\sqrt E }}$
$⇒$ $\log \lambda = \log \frac{h}{{\sqrt 2 m}} - \frac{1}{2}\log E$
$ \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.
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