Question

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

Options

• `lambda/3sqrt("M"/"m")`

• `lambda/3"M"/"m"`

• `lambda/3sqrt("m"/"M")`

• `lambda/3"m"/"M"`

Answer 1

An electron of mass m has de-Broglie wavelength λ when accelerated through potential difference V. When proton of mass M, is accelerated through potential difference 9V, the de-Broglie wavelength associated with it will be `underlinebb(lambda/3sqrt("m"/"M"))`. (Assume that wavelength is determined at low voltage)

Explanation:

From de-Broglie relation,

λ = `"h"/"p"` ⇒ λ = `"h"/sqrt(2"mKE")="h"/sqrt(2"mqV")`

⇒ λ ∝ `1/sqrt("qVm")`

For electron, λ_e ∝ `1/sqrt("eVm")`             ...(i)

For proton, λ_p = `1/sqrt("e9VM")`             ...(ii)

where, e is the charge on proton, potential difference = 9V and Mass of proton = m

From eqs. (i) and (ii)

`(lambda_"e")/(lambda"p") = sqrt((9"VMe")/"eVm")` 

⇒ λ_p = `lambda_e/3sqrt("m"/"M")`