QJEEM20Phc645: JEE Main 2020 Physics - Class 12 JEEM20Phc645 Solution: Particle A of mass m A = moving along the x-axis with velocity…

Question

Verified Solution • 2025–26 Session moderate

Particle A of mass m_A =

{m \over 2}

moving along the x-axis with velocity v₀ collides elastically with another particle B at rest having mass m_B =

{m \over 3}

. If both particles move along the x-axis after the collision, the change

\Delta

\lambda

in de-Broglie wavlength of particle A, in terms of its de-Broglie wavelength (

\lambda

₀) before collision is :

Accepted Answer

Applying momentum conservation

{m \over 2} \times {V_0} + {m \over 3} \times 0 = {m \over 2}{V_A} + {m \over 3}{V_B}

\Rightarrow

{{{V_0}} \over 2} = {{{V_A}} \over 2} + {{{V_B}} \over 3}

.....(1)

Since, collision is elastic (e = 1)

e = 1 =

{{{V_B} - {V_A}} \over {{V_0}}}

\Rightarrow

V₀ = V_B – V_A .....(2)

On solving (1) & (2) :

V_A =

{{{V_0}} \over 5}

Now, De-Broglie wavelength of A before collision :

\lambda

₀ =

{h \over {{m_A}{V_0}}}

=

{h \over {\left( {{m \over 2}} \right){V_0}}}

=

{{2h} \over {m{V_0}}}

Final De-Broglie wavelength :

\lambda

_f =

{h \over {{m_A}{V_A}}}

=

{h \over {\left( {{m \over 2}} \right)\left( {{{{V_0}} \over 5}} \right)}}

=

{{10h} \over {m{V_0}}}

Now,

\Delta

\lambda

=

\lambda

_f -

\lambda

₀

=

{{10h} \over {m{V_0}}}

-

{{2h} \over {m{V_0}}}

\Rightarrow

\Delta

\lambda

=

{{8h} \over {m{V_0}}}

\Rightarrow

\Delta

\lambda

=

4 \times {{2h} \over {m{V_0}}}

\Rightarrow

\Delta

\lambda

= 4

\lambda

₀

Step-by-Step Verified Solution