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An electron and an alpha particle have same kinetic energy. How are the de Broglie wavelengths associated with them related ? |
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Answer» SOLUTION :Kinetic ENERGY of the particle `K=1/2mv^(2)=(P^(2))/(2M)=(h^(2))/(2mlambda^(2))[lambda=h/P]` i.e `lambda =h/(sqrt(2mK)),lambda PROP1/(sqrtm)` `(lambda_e)/(lambda_prop)=sqrt((m_prop)/(m_e))` |
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