An electron (mass m) with an initial velocity $v=v_0\hat i$ is in an electric field $E=E_0\hat j$. If $λ_0h/mv_0$, its de- Broglie wavelength at time t is given by:
Answer & explanation
Correct answer: option 3
Initial de-Broglie wavelength of electron, $λ_0=\frac{h}{mv_0}$
Force on electron in electric field, $F = −eE = −eE_0\hat j$
Acceleration of electron, $a=\frac{F}{m}=\frac{eE_0\hat j}{m}$
It is acting along negative Y-axis.
The initial velocity of electron along X-axis, $v_{x0}=v_0\hat i$. Initial velocity of electron Y-axis, $v_{y0}= 0$
Velocity of electron after time t along X-axis, $v_x=v_0\hat i$ (∵ there is not electron along X-axis)
Velocity of electron after time t along Y-axis, $v_y=0+(-\frac{eE_0}{m}\hat j)t=-\frac{eE_0}{m}t\hat j$
Magnitude of velocity of electron after time t is $v=\sqrt{v_x^2+v_y^2}=\sqrt{v_0^2+(\frac{-eE_0}{m}t)^2}$
$⇒=v_0\sqrt{1+\frac{e^2E_0^2t^2}{m^2v_0^2}}$
de-Broglie wavelength, $λ'=\frac{h}{mv}$
$⇒=\frac{h}{mv_0\sqrt{1+e^2E_0^2t^2/(m^2v_0^2)}}=\frac{λ_0}{\sqrt{1+e^2E_0^2t^2/m^2v_0^2}}$