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| 1. |
An electron and alpha particle have the same de-Broglie wavelength associated with them. How are their kinetic energies related to each other ? |
| Answer» SOLUTION :As `LAMDA=(h)/(sqrt(2mK)) and lamda_(e)=lamda_(alpha)`, hence, we CONCLUDE that `(K_(e))/(K_(alpha))=(m_(alpha))/(m_(e))`. | |