An electron (mass m) with an initial velocity v=v_0 i is in an electric field E=E_0 j . If _0= h mv_0 , it’s de Breoglie wavelength at time t is given by:

An electron (mass m) with an initial velocity v=v_0 i is in an el…

MCQ

An electron $($mass m$)$ with an initial velocity $\text{v}=\text{v}_0\hat{\text{i}}$ is in an electric field $\text{E}=\text{E}_0\hat{\text{j}}$. If $\lambda_0=\frac{\text{h}}{\text{mv}_0}$, it’s de Breoglie wavelength at time t is given by:

A
$\lambda_0$
B
$\lambda_0\sqrt{1+\frac{\text{e}^2\text{E}_0^2\text{t}^2}{\text{m}^2\text{v}_0^2}}$
✓
$\frac{\lambda_0}{\sqrt{1+\frac{\text{e}^2\text{E}_0^2\text{t}^2}{\text{m}^2\text{v}_0^2}}}$
D
$\frac{\lambda_0}{\bigg({1+\frac{\text{e}^2\text{E}_0^2\text{t}^2}{\text{m}^2\text{v}_0^2}}\bigg)}$

✓

Answer

Correct option: C.

$\frac{\lambda_0}{\sqrt{1+\frac{\text{e}^2\text{E}_0^2\text{t}^2}{\text{m}^2\text{v}_0^2}}}$

According to the problem de$-$Broglie wavelength of electron at time $t = 0.$
is $\lambda_0=\frac{\text{h}}{\text{mv}_0}$
Electrostatic force on electron in electric field is
$\vec{\text{F}}_\text{e}=-\text{e}\vec{\text{E}}=-\text{e}\text{E}_0\hat{\text{j}}$
The acceleration of electron, $\vec{\text{a}}=\frac{\vec{\text{F}}}{\text{m}}=-\frac{\text{eE}_0}{\text{m}}\hat{\text{j}}$
It is acting along negative $y-$axis.
The intial velocity of electron along x-axis $\text{v}_{\text{x}_0}=\text{v}_0\hat{\text{i}}$.
This component of velocity will remain constant as there is no force on electron in this direction.
New considering $y-$diraction. Initial velocity of electron along $y-$axis, $\text{v}_{\text{y}_0}=0$.
Velocity of electron after time $t$ along $y-$axis,
$\text{v}_\text{y}=0+\Big(-\frac{\text{eE}_0}{\text{m}}\hat{\text{j}}\Big)\text{t}=-\frac{\text{eE}_0}{\text{m}}\text{t}\hat{\text{j}}$
Magnitude of velocity of electron after time t is
$\text{v}=\sqrt{\text{v}^2_\text{x}+\text{v}^2_\text{y}}=\sqrt{\text{v}_0^2+\Big(\frac{-\text{eE}_0}{\text{t}}\text{t}\Big)}^2$
$\Rightarrow\ =\text{v}_0\sqrt{1+\frac{\text{e}^2\text{E}_0^2\text{t}^2}{\text{m}^2\text{v}_0^2}}$
de$-$Broglie wavelength, $\lambda'=\frac{\text{h}}{\text{mv}}$
$=\frac{\text{h}}{\text{mv}_0\sqrt{1+\frac{\text{e}^2\text{E}_0^2\text{t}^2}{\text{m}^2\text{v}_0^2}}}=\frac{\lambda_0}{\sqrt{1+\frac{\text{e}^2\text{E}_0^2\text{t}^2}{\text{m}^2\text{v}_0^2}}}$
$\Rightarrow\ \lambda'=\frac{\lambda_0}{\bigg({1+\frac{\text{e}^2\text{E}_0^2\text{t}^2}{\text{m}^2\text{v}_0^2}}\bigg)}$