A particle of mass m kg and charge q coulomb is accelerated through V volt then the de-Broglie wavelength associated with it,in meter is `lambda`=…..m.

Options

A. `lambda=(h)/(mv)`
B. `lambda=(h)/(sqrt(2mqV))`
C. `lambda=(h)/(sqrt(mqV))`
D. `lambda=(hq)/(sqrt(2mV))`

Answer

B

Explanation

de-Broglie wavelength `lambda=(h)/(p)`….(1)
`(p^(2))/(2m)=qVimpliesp=sqrt(2Vqm)`……..(2)
putting vallue of equation (2) in equation (1),
`therefore lambda=(h)/(sqrt(2Vqm))`