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

Options

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

Answer

C

Explanation

The de-Brogile wavelengths and kinetic energy of a free electron are related as
`lambda=h/sqrt(2mK)implies lambda_(2)/lambda_(1)=sqrt(K_(1)/K_(2))=sqrt(1/2)implies lambda_(2)=lambda_(1)/sqrt(2)`