(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.

Explanation

(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.