Relativistic corrections become necessary when the expresion for the kinetic energy `1/2mv^(2)`, becomes comparable with `mc^(2)`, where m is the mass of the particle. At what de-broglie wavelength will relativistic corrections become important for an electron?

Options

A. `lambda=10nm`
B. `lambda=10^(-1)nm`
C. `lambda=10^(-4)nm`
D. `lambda=10^(-6)nm`

Answer

c,d

Explanation

Velocity of electron, `v=h//(mlambda)`
Let `h=6.6xx10^(-34)Js and m=9xx10^(-31)kg`
(a) When `lambda_(1)=10nm=10xx10^(-9)m=10^(-8)m`
`v_(1)=(6.6xx10^(-34))/((9xx10^(-31))xx10^(-8))=2.2/3xx10^(5)~~10^(5)m//s`
(b) When `lambda_(2)=10^(-1) nm=10^(-1)xx10^(-9)m=10^(-10)m`
`v_(2)=(6.6xx10^(-34))/((9xx10^(-31))xx10^(-10))~~10^(7)m//s`
(c) When `lambda_(3)=10^(-4) nm=10^(-4)xx10^(-9)m=10^(-13)m`
`v_(3)=(6.6xx10^(-34))/((9xx10^(-31))xx10^(-13))~~10^(10)m//s`
(d) When `lambda_(4)=10^(-6) nm=10^(-6)xx10^(-9)m=10^(-15)m`
`v_(4)=(6.6xx10^(-34))/((9xx10^(-31))xx10^(-15))~~10^(12)m//s`
As `v_(3) and v_(4)` are greater than velocity of light `(=3xx10^(8)m//s)`, hence relativistic correction is needed for `lambda=10^(-4)nm and lambda=10^(-6)nm`. thus, option (c) and (d) are required wavelength.