A particle of mass 1mg has the same wavelength as an electron moving with a velocity of `3xx10^6 ms^(-1)` . What is the velocity of the particle ?
(Take mass of the electron `=9xx10^(-31) kg` )

Options

A. `2.7xx10^(-21) m//s`
B. `2.7xx10^(-18) m//s`
C. `3.5xx10^(-17) m//s`
D. `9xx10^(-5) m//s`

Answer

B

Explanation

de broglie wavelength associated with the particle is given by `lamda= (h)/(mv) " and for the electron " lamda_e = (h)/(m_ev_e)`
But `lamda=lamda_e ("given") therefore (h)/(mv) = (h)/(m_e v_e) " or " mv = m_e v_e`
`therefore v = (m_ev_e)/(m) = (9xx10^(-31)xx3xx10^6)/(1xx10^(-6))`
`= 27xx10^(-19) = 2.7xx10^(-18) m//s`