A particle of mass 1 mg has the same wavelength as an electron moving with a velocity of `3xx10^(6)ms^(-1)`. The velocity of the particle is (mass of electron `=9.1xx10^(-31)kg`)

Options

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

Answer

B

Explanation

As per question `lamda=(h)/(mv)=(h)/(m_(e)v_(e))`
`implies v=(m_(e)v_(e))/(m)=((9.1xx10^(-31)kg)(3xx10^(6)ms^(-1)))/((1mg))=(9.1xx10^(-31)xx3xx10^(6))/(1xx10^(-6))=2.7xx10^(-18)ms^(-1)`.