Question 213: Dual Nature of Matter and Radiation

Question

Two electrons are moving with non-relativistic speed perpendicular to each other. If corresponding de Broglie wavelength are $${\lambda _1}$$ and $${\lambda _2},$$ their de Broglie wavelength in the frame of reference attached to their center of masses :

Accepted Answer

As we know,
momentum (p) = $${h \over \lambda }$$
Let  one perticle is moving in x direction and  other is in y dirrection.
$$	herefore\,\,\,\,$$ momentum of each electrons $${h \over {{\lambda _1}}}\widehat i$$ and $${h \over {{\lambda _2}}}\widehat j$$
$$	herefore\,\,\,\,$$  Velocity of each electrons $${h \over {m{\lambda _1}}}\widehat i$$ and $${h \over {m{\lambda _2}}}\widehat j$$
$$	herefore\,\,\,\,$$ Velocity of center of mass $$\left( {{{\overrightarrow \upsilon }_{_{CM}}}} ight)$$ = $${{\upsilon _1^{\widehat i} + \upsilon _2^{\widehat j}} \over 2}$$
Now, velocity of first electron about center of mass,
$${\overrightarrow \upsilon  _{_{1CM}}}$$ = $$\upsilon _1^{\widehat i}$$ $$-$$ $$\left( {{{\upsilon _1^{\widehat i} + \upsilon _2^{\widehat j}} \over 2}} ight)$$
= $${{\upsilon _1^{\widehat i} - \upsilon _2^{\widehat j}} \over 2}$$
Similarly,
$${\overrightarrow \upsilon  _{_{2CM}}} = {{\upsilon _2^{\widehat j} - \upsilon _1^{\widehat i}} \over 2}$$
Here,
$$\left| {{{\overrightarrow \upsilon  }_{_{1CM}}}} ight| = \left| {{{\overrightarrow \upsilon  }_{_{2CM}}}} ight| = {1 \over 2}$$
$$	herefore\,\,\,\,$$ $$\upsilon $$ = $${1 \over 2}$$ $$\sqrt {{{{h^2}} \over {{m^2}\lambda _1^2}} + {{{h^2}} \over {{m^2}\lambda _2^2}}} $$
$$ \Rightarrow $$$$\,\,\,\,$$ m  $$\upsilon $$ = $${1 \over 2}$$ $${\sqrt {{{{h^2}} \over {\lambda _1^2}} + {{{h^2}} \over {\lambda _2^2}}} }$$
$$ \Rightarrow $$$$\,\,\,\,$$ $${h \over {{\lambda _{CM}}}}$$ = h$${\sqrt {{1 \over {4\lambda _1^2}} + {1 \over {4\lambda _2^2}}} }$$
$$ \Rightarrow $$$$\,\,\,\,$$ $${1 \over {{\lambda _{CM}}}} = {{\sqrt {\lambda _1^2 + \lambda _2^2} } \over {2{\lambda _1}{\lambda _2}}}$$
$$ \Rightarrow $$$$\,\,\,\,$$  $${\lambda _{CM}} = {{2{\lambda _1}{\lambda _2}} \over {\sqrt {\lambda _1^2 + \lambda _2^2} }}$$

Answer

$${\lambda _{CM}} = {\lambda _1} = {\lambda _2}$$

Answer

$${\lambda _{CM}} = {{2{\lambda _1}{\lambda _2}} \over {\sqrt {\lambda _1^2 + \lambda _2^2} }}$$

Answer

$${1 \over {{\lambda _{CM}}}} = {1 \over {{\lambda _1}}} + {1 \over {{\lambda _2}}}$$

Answer

$${\lambda _{CM}} = \left( {{{{\lambda _1} + {\lambda _2}} \over 2}} \right)$$

Dual Nature of Matter and Radiation

Two electrons are moving with non-relativistic speed perpendicular to each other. If corresponding de Broglie wavelength are ${\lambda _1} and {\lambda _2},$ their de Broglie wavelength in the frame of reference attached to their center of masses :

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