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(a) Draw a graph showing the variation of de-Broglie wavelength of a particle of charge 'q' and mass 'm' with accelerating potential 'V'. (b) Proton and deuteron have the same de-Broglie wavelength. Explain which has more kinetic enegy. |
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Answer» Solution :(B) We know that de-Broglie WAVELENGTH `LAMDA=(h)/(sqrt(2mK))` As per QUESTION `lamda_("proton")=lamda_("deuteron")`, hence we have `m_(p)K_(p)=m_(d)K_(d)implies K_(d)=(m_(p))/(m_(d))*K_(p)` Since mass `(m_(d))` of deuteron particle is more than that of proton `(m_(p)),`, hence `K_(d) lt K_(p)`. so the proton has more kinetic energy. |
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