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.

Explanation

Given, `v_("particle")=3 v_("electron")` …(i)
and `lamda_("particle")=1.813xx10^(-4)lamda_("electron")`
(i) As, `lamda=h/(mv)` [de-Broglie equation]
`rArr(m_("particle"))/(m_("electron"))=(lamda_("electron")xxv_("electron"))/(lamda_("particle")xxv_("electron"))`
`=(lamda_("electron")xxv_("electron"))/(1.813xx10^(-4)xxlamda_("electron")xx3v_("electron"))`
`therefore m_("particle")=1839 m_("electron")` [from Eq. (i)]
`m_("particle")=1.839xx9.1xx10^(-31)`
=`1.673xx10^(-27) kg`
Particle is either a proton or a neutron.