Question 191: 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: