A particle of charge q, mass m and energy E has de-Broglie wavelength `lambda`. For a particle of charge 2q, mass 2m and energy 2E, the de-Broglie wavelength is

Options

A. `lambda/4`
B. `2 lambda`
C. `8 lambda`
D. `lambda/2`

Answer

D

Explanation

The de-Broglie wavelength `lambda` of charged particle of charge q mass m and energy E is given by
`lambda =h/sqrt(2mE)` ...(i)
For another particle mass `m'=2 m`
charge `q'=2q`
and energy, `E'=2E`
de-Broglie wavelength,
`lambda'=h/sqrt(2m'E')`
`=h/sqrt(2xx2mxx2E)`
`=1/2. h/sqrt(2 m E) lambda'=lambda/2`