de Broglie wavelength associated with an electron acclerated through a potential difference `V` is `lambda`. What will be its wavelength when the accelerating potential is increased to `4 V`?

Explanation

de Broglie wavelength, `lambda prop (1)/(sqrt(V))`
`:. (lambda_(1))/(lambda_(2))=sqrt((V_(2))/(V_(1)))`
`:. (lambda_(1))/(lambda_(2))sqrt((4V)/(V))=2 rArr lambda_(2)(lambda_(1))/(2)`