The de - Broglie wavelength of a particle accelerated with `150 volt` "potential is" `10^(-10) m`. If it is accelerated by `600 volts p.d`., its wavelength will be

Options

A. `0.25 Å`
B. `0.5 Å`
C. `1.5 Å`
D. `2 Å`

Answer

B

Explanation

By using `lambda prop (1)/(sqrt(V)) rArr (lambda_(1))/(lambda_(2)) = sqrt((V_(2))/(V_(1)))`
`rArr (10^(-10))/(lambda_(2)) = sqrt((600)/(150)) = 2 rArr lambda_(2) = 0.5 Å`.