Question

Two particles $A$ and $B$ are in motion. If the wavelength associated with particle $A$ is $5\times {10}^{-8}$m. Calculated the wavelength (in $\mathring{A}$) associated with particle $B$ if its momentum is half of $A$.

Answer 1

• One wavelength equals the distance between two successive wave crests or troughs.
  
• We know De-Broglies wavelength is given by,

$\lambda =\frac{h}{p}$, where h is Planck's constant and p is momentum

Given,

${\lambda }_A=5\times {10}^{-8}m$ and momentum of B is half of momentum of A.

So,

$⟹\frac{{\lambda }_A}{{\lambda }_B}=0.5$

$⟹{\lambda }_B=2\times {\lambda }_A$

$⟹{\lambda }_B=10\times {10}^{-8}=1000Angstrom$