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

de-Broglie wavelength associated with electron moving with velocity v,
`lambda=(h)/(mv)`
So, `lambda_(e)=(h)/(9.1xx10^(-31)xx3xx10^(6))`
Wavelength of particle of mass 1 mg moving with velocity v.
`lambda_(p)=(h)/(10^(-3)xxv)`
As given, `lambda_(e)=lambda_(p)`
`rArr (h)/(10^(-3)xxv)=(h)/(9.1xx10^(-31)xx3xx10^(6))`
`v=(27.3xx10^(25))/(10^(-3))m//s=2.73xx10^(-21)m//s`