Relativistic corrections become necessary when the expression 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. `lamda=1nm`
B. `lamda=10nm`
C. `lamda=10^(-1)`
D. `lamda=10^(-4)`

Answer

D

Explanation

(d) : Here h `=6.63xx10^(-34)Js,m=9xx10^(-31)kg`
Velocity of electron,
`v=(h)/(mlamda)` . . . (i)
(i) `"When"lamda_(1)=1nm=10^(-9)m`
`v_(1)=(6.63xx10^(-34))/(9xx10^(-31)xx10^(-9))=0.74xx10^(6)" m s"^(-1)`
(ii) `"When"lamda_(2)=10nm=10xx10^(-9)m=10^(-8)m`
`v_(2)=(6.63xx10^(-34))/(9xx10^(-31)xx10^(-8))=10^(5)"m s"^(-1)`
(iii) `"When"lamda_(3)=10^(-1)nm=10^(-1)xx10^(-9)m=10^(-10)m`
`v_(3)=(6.63xx10^(-34))/(9xx10^(-31)xx10^(-10))=10^(7)"m s"^(-1)`
(iv) `"When"lamda_(4)=10^(-4)nm=10^(-4)xx10^(-9)m=10^(-13)m`
`v_(4)=(6.63xx10^(-34))/(9xx10^(-31)xx10^(-13))=10^(10)"m s"^(-1)`
As `v_(4)` is greater than the velocity of light `(=3xx10^(8)" m s"^(-1))` hence correction is needed for `lamda=10^(-4)nm`.