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`.

Options

A. `1.67xx10^(-27)kg`
B. `1.67xx10^(-31)kg`
C. `1.67xx10^(-19)kg`
D. `1.67xx10^(-14)kg`

Answer

A

Explanation

(a) : de Broglie wavelength of a moving jparticle having mass m and velocity v is given by
`lamda=(h)/(p)=(h)/(mv) or m=(h)/(lamdav)`
For an electron `lamda_(e)=(h)/(m_(e)v_(e))orm_(e)=(h)/(lamda_(e)v_(e))`
Given `(v)/(v_(e))=3and(lamda)/(lamda_(e))=1.813xx10^(-4)`
Mass of the particle, `m=m_(e)((lamda_(e))/(lamda))((v_(e))/(v))`
Substituting the value, we get
`m=9.1xx10^(-31)xx(1)/(1.813xx10^(-4))xx(1)/(3)`
or `m=1.67xx10^(-27)kg`