De-broglie wavelength associated with an electron accelerated through a potential difference `4V` is `lambda_(1)`.When accelerating voltage is decreased by `3V`, its de-broglie wavelength `lambda_(2)`. The ratio `lambda_(2)//lambda_(1)`is :

Explanation

De-broglie wavelength,
`lambda=h/(sqrt(2mqV))` or `lambda prop 1/(sqrt(V))`
`:. (lambda_(2))/(lambda_(1))=sqrt((V_(1))/(V_(2)))=sqrt((4V)/(4V-3V))=2`