A particle of mass `1 mg` has the same wavelength as an electron moving with a velocity of `3 xx 10^(6) ms^(-1)`. The velocity of the particle is

Options

A. `3xx10^(-31)ms^(-1)`
B. `2.7xx10^(-21)ms^(-1)`
C. `2.7xx10^(-18)ms^(-1)`
D. `9xx10^(-2)ms^(-1)`

Answer

C

Explanation

We know that `p=(h)/(lamda)` or `mv=(h)/(lamda)`
`therefore`As both particle and electrons having same wavelength therefore, their momentum will be equal to
`impliesm_(p)v_(p)=m_(e)v_(e)`
`implies v_(p)=(m_(e)v_(e))/(m_(p))`
`=(9.1xx10^(-31)xx3xx10^(6))/(10^(-6))`
`implies v_(p)=2.7xx10^(-18)ms^(-1)`