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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`)


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