An electron of mass m has de broglie wavelength `lamda` when accelerated through a potential difference V . When a proton of mass M is accelerated through a potential difference 9V, the de broglie wavelength associated with it will be (Assume that wavelength is determined at low voltage ) .

Options

A. a. `lamda/3 sqrt(M/m)`
B. b.`lamda/3 . M/m`
C. c.`lamda/3 sqrt(m/M)`
D. d.`lamda/3 . m/M`

Answer

C

Explanation

For a charged particle accelerated through a P.D. (V ),
`K.E. = qV " and " lamda= (h)/(sqrt(2mE)) = (h)/(sqrt(2mqV))`
`therefore lamda_2/lamda_1 = sqrt((2mqV)/(2Mq(9V))`
the electron and the proton have the same charge (numerically).
`therefore lamda_2/lamda_1 = 1/3 sqrt(m/M)`
or `lamda_p = (lamda)/(3) sqrt(m/M) [ therefore lamda_1 = lamda " and" lamda_2 = lamda_p]`