Which of the following figures represent the variation of particle momentum and the associated de-Broglie wavelength?

• Correct answer
  A

  
• B

  
• C

  
• D

  

Solution

de-Broglie's wavelength, $\mathrm{\lambda }=\frac{\mathrm{h}}{\mathrm{p}}$
$\Rightarrow \mathrm{p}\mathrm{\lambda }=\mathrm{h}$
$\Rightarrow \mathrm{\lambda }\mathrm{p}=\mathrm{h}$, Same as $\left(\mathrm{x}\mathrm{y}=\mathrm{c}\right)$
Momentum, $\mathrm{p}\propto \frac{1}{\mathrm{\lambda }}$ (Rectangular hyperbola)

Hence, the correct answer is option: 'A'.