An electron of mass m and magnitude of charge e at rest, gets accelerated by a constant electric field E . The rate of change of de-Broglie wavelength of this electron at a time t is (ignore relativistic effects)

Question

An electron of mass m and magnitude of charge e at rest, gets accelerated by a constant electric field E . The rate of change of de-Broglie wavelength of this electron at a time t is (ignore relativistic effects) (A) (d λ /d t)=-(h/e E t) (B) (d λ /d t)=-(2 h/e E t) (C) (d λ /d t)=-(2 h/e E t2) (D) (d λ /d t)=-(h/e E t2)

Tardigrade Admin · Accepted Answer

The acceleration of the electron is a=(e E/m) The velocity of the electron as a function of time is v=at=(e E/m)t The de-Broglie wavelength is λ =(h/m v) λ =(h/m ((e E)m) t) λ =(h/e E t) (d λ /d t)=-(h/e E t2)

Q. An electron of mass $m$ and magnitude of charge $e$ at rest, gets accelerated by a constant electric field $E$ . The rate of change of de-Broglie wavelength of this electron at a time $t$ is (ignore relativistic effects)

$\frac{d λ}{d t} = - \frac{h}{e Et}$

$\frac{d λ}{d t} = - \frac{2 h}{e Et}$

$\frac{d λ}{d t} = - \frac{2 h}{e E t ^{2}}$

$\frac{d λ}{d t} = - \frac{h}{e E t ^{2}}$

The acceleration of the electron is
$a = \frac{e E}{m}$
The velocity of the electron as a function of time is
$v = a t = \frac{e E}{m} t$
The de-Broglie wavelength is
$λ = \frac{h}{m v}$
$λ = \frac{h}{m ( \frac{e E}{m} ) t}$
$λ = \frac{h}{e Et}$
$\frac{d λ}{d t} = - \frac{h}{e E t ^{2}}$

Q. An electron of mass m and magnitude of charge e at rest, gets accelerated by a constant electric field E . The rate of change of de-Broglie wavelength of this electron at a time t is (ignore relativistic effects)

dtdλ​=−eEth​

dtdλ​=−eEt2h​

dtdλ​=−eEt22h​