Question 170: Dual Nature of Matter and Radiation

Question

An electron (mass m) with initial velocity $\overrightarrow v  = {v_0}\widehat i + {v_0}\widehat j$ is in an electric field $\overrightarrow E  =  - {E_0}\widehat k$. If $\lambda _0$ is initial de-Broglie wavelength of electron,
its de-Broglie wave length at time t is given
by :

Accepted Answer

$\overrightarrow v  = {v_0}\widehat i + {v_0}\widehat j$
$\lambda $
0
= ${h \over {m\sqrt 2 {v_0}}}$ ....(1)
$\overrightarrow E  =  - {E_0}\widehat k$
$\overrightarrow F  = q\overrightarrow E $ = (-e)($- {E_0}\widehat k$) = $e{E_0}\widehat k$
$ 	herefore $ $\overrightarrow a  = {{\overrightarrow F } \over m}$ = ${{e{E_0}\widehat k} \over m}$
Velocity at time t,
$\overrightarrow {{v_f}} $ = ${v_0}\widehat i + {v_0}\widehat j$ + ${{e{E_0}} \over m}t\widehat k$
$\left| {\overrightarrow {{v_f}} } ight|$ = $\sqrt {v_0^2 + v_0^2 + {{\left( {{{e{E_0}t} \over m}} ight)}^2}} $
= $\sqrt {2v_0^2 + {{{e^2}E_0^2} \over {{m^2}}}{t^2}} $
$ 	herefore $ Wavelength at time t
$\lambda $ = ${h \over {m{v_f}}}$ = ${h \over {m\sqrt {2v_0^2 + {{{e^2}E_0^2} \over {{m^2}}}{t^2}} }}$
= ${h \over {\sqrt 2 m{v_0}\sqrt {1 + {{{e^2}E_0^2} \over {2{m^2}v_0^2}}{t^2}} }}$
= ${{{\lambda _0}} \over {\sqrt {1 + {{{e^2}E_0^2} \over {2{m^2}v_0^2}}{t^2}} }}$

Answer

${{{\lambda _0} } \over {\sqrt {1 + {{{e^2}{E^2}{t^2}} \over {{m^2}v_0^2}}} }}$

Answer

${{{\lambda _0}\sqrt 2 } \over {\sqrt {1 + {{{e^2}{E^2}{t^2}} \over {{m^2}v_0^2}}} }}$

Answer

${{{\lambda _0} } \over {\sqrt {1 + {{{e^2}{E^2}{t^2}} \over {2{m^2}v_0^2}}} }}$

Answer

${{{\lambda _0}} \over {\sqrt {2 + {{{e^2}{E^2}{t^2}} \over {{m^2}v_0^2}}} }}$

Dual Nature of Matter and Radiation

An electron (mass m) with initial velocity \overrightarrow v = {v_0}\widehat i + {v_0}\widehat j is in an electric field \overrightarrow E = - {E_0}\widehat k. If \lambda _0 is initial de-Broglie wavelength of electron, its de-Broglie wave length at time t is given by :

Choose the correct answer: