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 identity the particle. Mass of electron `=9.11xx10^(-31)kg`.

Explanation

`(lambda_p)/(lambda_e)=(h//(m_pv_p))/(h//(m_ev_e))=(m_ev_e)/(m_pv_p)=(m_e)/(m_p)xx1/3`
or `m_p=(m_e)/(3(lambda_p//lambda_e))=(9.11xx10^(-31))/(3xx1.813xx10^-4)`
`=1.675xx10^(-27)kg`
It is the mass of a neutron or proton. Hence the emitted particle is a neutron or proton.