If kinetic energy of emitted electron is made double then its de-Broglie wavelength will become……times initial de-Broglie wavelength.

Options

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

Answer

C

Explanation

de-Broglie wavelength `lambda=(h)/(p)`
but `p^(2)=sqrt(2mK)`
`therefore lambda =(h)/(sqrt(2mK))`
`therefore lambda prop (1)/(sqrt(K))` [`because(h)/(sqrt(2m))` same ]
`therefore (lambda_(2))/(lambda_(1))=sqrt((K_(1))/(K_(2)))`
`therefore (lambda_(2))/(lambda_(1))=sqrt((K)/(2K)) therefore (lambda_(2))/(lambda_(1))=(1)/(sqrt(2))`