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

The particle is electron as for electron accelerated with
`150 V lambda = sqrt((150)/(150))Å = 10^(-10)m`
`therefore` For 600 V
`lambda = sqrt((150)/(600)) Å = 0.5 Å`