Saved Bookmarks
| 1. |
What is the de Broglie wavelength associated with a particle having momentum 10-26 kg∙m/s? [h = 6.63 × 10-34 J∙s] |
|
Answer» The de Broglie wavelength associated with the particle, λ = \(\cfrac hp\) = \(\cfrac{6.63\times10^{-34}}{10^{-26}}\) = 6.63 × 10-8 m |
|