JEE MAIN - Physics (2019 - 9th April Evening Slot - No. 19)
A particle 'P' is formed due to a completely
inelastic collision of particles 'x' and 'y' having
de-Broglie wavelengths '$$\lambda $$x' and '$$\lambda $$y'
respectively. If x and y were moving in opposite
directions, then the de-Broglie wavelength of
'P' is :-
$${\lambda _x} - {\lambda _y}$$
$${{{\lambda _x}{\lambda _y}} \over {\left| {{\lambda _x} - {\lambda _y}} \right|}}$$
$${\lambda _x} + {\lambda _y}$$
$${{{\lambda _x}{\lambda _y}} \over {{\lambda _x} + {\lambda _y}}}$$
Explanation
Conservation of momentum
$$\overrightarrow {{p_x}} + \overrightarrow {{p_y}} = {\overrightarrow p _{final}}$$
mxvx – myvy = (mx + my) V
$${h \over {{\lambda _x}}} - {h \over {{\lambda _y}}} = {h \over \lambda }$$
$$ \Rightarrow \lambda = {{{\lambda _x}{\lambda _y}} \over {\left| {{\lambda _x} - {\lambda _y}} \right|}}$$
$$\overrightarrow {{p_x}} + \overrightarrow {{p_y}} = {\overrightarrow p _{final}}$$
mxvx – myvy = (mx + my) V
$${h \over {{\lambda _x}}} - {h \over {{\lambda _y}}} = {h \over \lambda }$$
$$ \Rightarrow \lambda = {{{\lambda _x}{\lambda _y}} \over {\left| {{\lambda _x} - {\lambda _y}} \right|}}$$
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