`lambda_(e),lambda_(p)` and `lambda_(alpha)` are the de-Broglie wavelength of electron, proton and `alpha` particle. If all the accelerated by same potential, then

Options

A. `lambda_(e) lt lambda_(p)lt lambda_(alpha)`
B. `lambda_(e) lt lambda_(p)gt lambda_(alpha)`
C. `lambda_(e) gt lambda_(p)lt lambda_(alpha)`
D. `lambda_(e) gt lambda_(p)gt lambda_(alpha)`

Answer

D

Explanation

de - Broglie wavelnegth can be givena as
`lambda=(h)/(p)=(h)/(sqrt(2mE)) or lambda prop (1)/(sqrt(Em))`
Now kinetic energy E gained by a charged particle under potentialy V is `E=qV`
given V is same for the given three particles
`therefore E_e =eV, E_p=eV`
`E_alpha=2eV rArr E_e = E_p lt E_alpha and m_e lt m_p lt m_alpha`
`lambda=(h)/(sqrt(2m_eE_e)) gt (h)/(sqrt(2m_pE_p) )gt (h)/(sqrt(2m_alpha Ealpha))lambda_egtlambda_pgtlambda_alpha`