de-Broglie proposed the dual nature of electron and put forward his wave concept. The wave length of electron in an orbit was given by `lambda =h/(mv)`
The de-Broglie.s wavelength `(lambda)` of the electron subjected to an accelerating potential of V volts is given by

Options

A. `(ch)/(sqrt(2mv))`
B. `(h)/(sqrt(2mev))`
C. `(h)/(sqrt(2me))xx v`
D. `(mh)/(sqrt(2eV))`

Answer

B

Explanation

`lambda = h/m sqrt((m)/(2eV)) = (h)/(sqrt(2meV))`