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| 1. |
An electron and a proton are possessing same amount of KE. Which of the two has greater de Broglie wavelength Explain. |
| Answer» Solution :The DE Brogile wavelength, `lambda=h/sqrt(2ME)` where E is KE of particle mass m. Here KE is same for PROTON and ELCETRON. So `lambda_alpha 1/sqrtm`.HENCE `lambda` will be greater for smaller mass. Thus de Broglie wavelength is greater for an electron than a proton (`lambda_e>lambda_p`) | |