An electron of mass me and a proton of mass mP are accelerated through the same potential difference. The ratio of the de-Broglie wavelength associated with the electron to that with the proton is :-

Question

An electron of mass me and a proton of mass mP are accelerated through the same potential difference. The ratio of the de-Broglie wavelength associated with the electron to that with the proton is :- (A) (mp/me) (B) 1 (C) √(mp/me) (D) (me/mp)

Tardigrade Admin · Accepted Answer

K E=e Δ V λe=(h/√2 me(e Δ V)) λp=(h/√2 mp(e Δ V)) ⇒ (λe/λp)=√(mp/me)

Q. An electron of mass $m_{e}$ and a proton of mass $m_{P}$ are accelerated through the same potential difference. The ratio of the de-Broglie wavelength associated with the electron to that with the proton is :-

$\frac{m _{p}}{m _{e}}$

$1$

$\frac{m _{p}}{m _{e}}$

$\frac{m _{e}}{m _{p}}$

$K E = e Δ V$
$λ_{e} = \frac{h}{2 m _{e} ( e Δ V )}$
$λ_{p} = \frac{h}{2 m _{p} ( e Δ V )}$
$\Rightarrow \frac{λ _{e}}{λ _{p}} = \frac{m _{p}}{m _{e}}$

Q. An electron of mass me​ and a proton of mass mP​ are accelerated through the same potential difference. The ratio of the de-Broglie wavelength associated with the electron to that with the proton is :-

me​mp​​

1

me​mp​​​

mp​me​​

KE=eΔV λe​=2me​(eΔV)​h​ λp​=2mp​(eΔV)​h​ ⇒λp​λe​​=me​mp​​​

Q. An electron of mass $m_{e}$ and a proton of mass $m_{P}$ are accelerated through the same potential difference. The ratio of the de-Broglie wavelength associated with the electron to that with the proton is :-

$\frac{m _{p}}{m _{e}}$

$1$

$\frac{m _{p}}{m _{e}}$

$\frac{m _{e}}{m _{p}}$

$K E = e Δ V$
$λ_{e} = \frac{h}{2 m _{e} ( e Δ V )}$
$λ_{p} = \frac{h}{2 m _{p} ( e Δ V )}$
$\Rightarrow \frac{λ _{e}}{λ _{p}} = \frac{m _{p}}{m _{e}}$