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

Explanation

Let `lamda_p,m_p and v_p` be de Broglie wavelength , mass and velocity of the particle, respectively . Then
`lamda_p=h/(m_pv_p)`
The `lamda_e,m_e and v_e` are de Broglie wavelength , mass and velocity of the electron , then
`lamda_e=h/(m_ev_e)`
`:. (lamda_p)/(lamda_e)=(m_ev_e)/(m_pv_p)implies(m_p)/(m_e)=(lamda_e)/(lamda_p).(v_e)/(v_p)`
Given, `v_p/v_e=3impliesv_e/v_p=1/3`
Also, `lamda_p/lamda_e=1.813xx10^(-4)`
or `lamda_e/lamda_p=1/(1.813xx10^(-4))`
`:.m_p/m_e=1/(1.813xx10^(-4))xx1/3`
`impliesm_p=m_e/(5.439xx10^(-4))=(9.11xx10^(-31))/(5.439xx10^(-4))`
`=1.675xx10^(-27)kg`
The particle with this mass could be a proton or a neutron.