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(a) A particle is moving three times as fast as an electron. The ratio of the de-Broglie wavelength of the particle to that of the electron is 1.813xx10^(-4). Calculate the particle's mass and identify the particle. (b) An electron and proton have the same kinetic energy. which of the two will have larger de Broglie wavelength ? Give reason. |
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Answer» Solution :(a) As per question `v_("particle")=3v_("electron") and lamda_("particle")=1.813xx10^(-4)lamda_("electron")`. From the relation `lamda=(h)/(mv)`, we have `(m_("particle"))/(m_("electron"))=(lamda_("electron")xxv_("electron"))/(lamda_("particle")xxv_("particle"))=(lamda_("electron")xxv_("electron"))/((1.813xx10^(-4)lamda_("electron"))xx(3v_("electron")))=1839` `implies m_("particle")=1839m_("electron")=1839xx9.1xx10^(-31)kg=1.673xx10^(-27)kg` It SHOWS that the given particle is either a PROTON or a neutron. (b) From the relation `lamda=(h)/(SQRT(2mK))`, it is clear that for same kinetic energy, de-Broglie wavelength of electron is more because its mass is less than that of proton. |
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