QJEEM18Phdf88: JEE Main 2018 Physics - Class 12 JEEM18Phdf88 Solution: Two electrons are moving with non-relativistic speed…

Question

Verified Solution • 2025–26 Session moderate

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 :

A ${\lambda _{CM}} = {\lambda _1} = {\lambda _2}$
B ${\lambda _{CM}} = {{2{\lambda _1}{\lambda _2}} \over {\sqrt {\lambda _1^2 + \lambda _2^2} }}$
C ${1 \over {{\lambda _{CM}}}} = {1 \over {{\lambda _1}}} + {1 \over {{\lambda _2}}}$
D ${\lambda _{CM}} = \left( {{{{\lambda _1} + {\lambda _2}} \over 2}} \right)$

Accepted Answer

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} }}

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