JEE MAIN - Physics (2018 - 15th April Morning Slot - No. 8)
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 :
$${\lambda _{CM}} = {\lambda _1} = {\lambda _2}$$
$${\lambda _{CM}} = {{2{\lambda _1}{\lambda _2}} \over {\sqrt {\lambda _1^2 + \lambda _2^2} }}$$
$${1 \over {{\lambda _{CM}}}} = {1 \over {{\lambda _1}}} + {1 \over {{\lambda _2}}}$$
$${\lambda _{CM}} = \left( {{{{\lambda _1} + {\lambda _2}} \over 2}} \right)$$
Explanation
As we know,
momentum (p) = $${h \over \lambda }$$
Let one perticle is moving in x direction and other is in y dirrection.
$$\therefore\,\,\,\,$$ momentum of each electrons $${h \over {{\lambda _1}}}\widehat i$$ and $${h \over {{\lambda _2}}}\widehat j$$
$$\therefore\,\,\,\,$$ Velocity of each electrons $${h \over {m{\lambda _1}}}\widehat i$$ and $${h \over {m{\lambda _2}}}\widehat j$$
$$\therefore\,\,\,\,$$ Velocity of center of mass $$\left( {{{\overrightarrow \upsilon }_{_{CM}}}} \right)$$ = $${{\upsilon _1^{\widehat i} + \upsilon _2^{\widehat j}} \over 2}$$
Now, velocity of first electron about center of mass,
$${\overrightarrow \upsilon _{_{1CM}}}$$ = $$\upsilon _1^{\widehat i}$$ $$-$$ $$\left( {{{\upsilon _1^{\widehat i} + \upsilon _2^{\widehat j}} \over 2}} \right)$$
= $${{\upsilon _1^{\widehat i} - \upsilon _2^{\widehat j}} \over 2}$$
Similarly,
$${\overrightarrow \upsilon _{_{2CM}}} = {{\upsilon _2^{\widehat j} - \upsilon _1^{\widehat i}} \over 2}$$
Here,
$$\left| {{{\overrightarrow \upsilon }_{_{1CM}}}} \right| = \left| {{{\overrightarrow \upsilon }_{_{2CM}}}} \right| = {1 \over 2}$$
$$\therefore\,\,\,\,$$ $$\upsilon $$ = $${1 \over 2}$$ $$\sqrt {{{{h^2}} \over {{m^2}\lambda _1^2}} + {{{h^2}} \over {{m^2}\lambda _2^2}}} $$
$$ \Rightarrow $$$$\,\,\,\,$$ m $$\upsilon $$ = $${1 \over 2}$$ $${\sqrt {{{{h^2}} \over {\lambda _1^2}} + {{{h^2}} \over {\lambda _2^2}}} }$$
$$ \Rightarrow $$$$\,\,\,\,$$ $${h \over {{\lambda _{CM}}}}$$ = h$${\sqrt {{1 \over {4\lambda _1^2}} + {1 \over {4\lambda _2^2}}} }$$
$$ \Rightarrow $$$$\,\,\,\,$$ $${1 \over {{\lambda _{CM}}}} = {{\sqrt {\lambda _1^2 + \lambda _2^2} } \over {2{\lambda _1}{\lambda _2}}}$$
$$ \Rightarrow $$$$\,\,\,\,$$ $${\lambda _{CM}} = {{2{\lambda _1}{\lambda _2}} \over {\sqrt {\lambda _1^2 + \lambda _2^2} }}$$
momentum (p) = $${h \over \lambda }$$
Let one perticle is moving in x direction and other is in y dirrection.
$$\therefore\,\,\,\,$$ momentum of each electrons $${h \over {{\lambda _1}}}\widehat i$$ and $${h \over {{\lambda _2}}}\widehat j$$
$$\therefore\,\,\,\,$$ Velocity of each electrons $${h \over {m{\lambda _1}}}\widehat i$$ and $${h \over {m{\lambda _2}}}\widehat j$$
$$\therefore\,\,\,\,$$ Velocity of center of mass $$\left( {{{\overrightarrow \upsilon }_{_{CM}}}} \right)$$ = $${{\upsilon _1^{\widehat i} + \upsilon _2^{\widehat j}} \over 2}$$
Now, velocity of first electron about center of mass,
$${\overrightarrow \upsilon _{_{1CM}}}$$ = $$\upsilon _1^{\widehat i}$$ $$-$$ $$\left( {{{\upsilon _1^{\widehat i} + \upsilon _2^{\widehat j}} \over 2}} \right)$$
= $${{\upsilon _1^{\widehat i} - \upsilon _2^{\widehat j}} \over 2}$$
Similarly,
$${\overrightarrow \upsilon _{_{2CM}}} = {{\upsilon _2^{\widehat j} - \upsilon _1^{\widehat i}} \over 2}$$
Here,
$$\left| {{{\overrightarrow \upsilon }_{_{1CM}}}} \right| = \left| {{{\overrightarrow \upsilon }_{_{2CM}}}} \right| = {1 \over 2}$$
$$\therefore\,\,\,\,$$ $$\upsilon $$ = $${1 \over 2}$$ $$\sqrt {{{{h^2}} \over {{m^2}\lambda _1^2}} + {{{h^2}} \over {{m^2}\lambda _2^2}}} $$
$$ \Rightarrow $$$$\,\,\,\,$$ m $$\upsilon $$ = $${1 \over 2}$$ $${\sqrt {{{{h^2}} \over {\lambda _1^2}} + {{{h^2}} \over {\lambda _2^2}}} }$$
$$ \Rightarrow $$$$\,\,\,\,$$ $${h \over {{\lambda _{CM}}}}$$ = h$${\sqrt {{1 \over {4\lambda _1^2}} + {1 \over {4\lambda _2^2}}} }$$
$$ \Rightarrow $$$$\,\,\,\,$$ $${1 \over {{\lambda _{CM}}}} = {{\sqrt {\lambda _1^2 + \lambda _2^2} } \over {2{\lambda _1}{\lambda _2}}}$$
$$ \Rightarrow $$$$\,\,\,\,$$ $${\lambda _{CM}} = {{2{\lambda _1}{\lambda _2}} \over {\sqrt {\lambda _1^2 + \lambda _2^2} }}$$
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