If the kinetic energy of a free electron doubles , its de - Broglie wavelength changes by the factor

Options

A. `(1)/(2)`
B. `2`
C. `(1)/(sqrt(2))`
D. `sqrt(2)`

Answer

C

Explanation

We know
`lambda=(h)/(m v)`
and `K=(1)/(2) mv^(2)=((3v)^(2))/(2m)`
`rArrmv=sqrt(2mK)`
Thus `lambda=(h)/(sqrt(2mK))`
`rArr lambda prop (1)/(sqrt(K))`
`:. (lambda_(2))/(lambda_(1))=sqrt(K_(1))/(sqrt(K_(2)))= sqrt(2K_(1))/(sqrt(2K_(1))) " "( :. K_(2)=2K_(1))`
`rArr (lambda_(2))/(lambda)=(1)/(sqrt(2))`
`:. lambda_(2)=(1)/(sqrt(2))`