MCQ
Two particles move at right angle to each other. Their de Broglie wavelengths are ${\lambda _1}$ and ${\lambda _2}$ respectively. The particles suffere perfectly inelastic collision. The de Broglie wavelength $\lambda $ , of the final particle, is given by
  • A
    $\lambda  = \sqrt {{\lambda _1}{\lambda _2}} $
  • B
    $\lambda  = \frac{{{\lambda _1} + {\lambda _2}}}{2}$
  • C
    $\frac{2}{\lambda } = \frac{1}{{{\lambda _1}}} + \frac{1}{{{\lambda _2}}}$
  • $\frac{1}{{{\lambda ^2}}} = \frac{1}{{\lambda _1^2}} + \frac{1}{{\lambda _2^2}}$

Answer

Correct option: D.
$\frac{1}{{{\lambda ^2}}} = \frac{1}{{\lambda _1^2}} + \frac{1}{{\lambda _2^2}}$
d
$\vec{P}_{1}=\frac{h}{\lambda_{1}} \hat{i}$ and $\vec{P}_{2}=\frac{h}{\lambda_{2}} \hat{j}$

Using momentum conservation

$\vec{P}=\vec{P}_{1}+\vec{P}_{2}=\frac{h}{\lambda_{1}} \hat{i}+\frac{h}{\lambda_{2}} \hat{j}$

$|\vec P| = \sqrt {{{\left( {\frac{h}{{{\lambda _1}}}} \right)}^2} + {{\left( {\frac{h}{{{\lambda _2}}}} \right)}^2}} $

$\frac{h}{\lambda}=\sqrt{\left(\frac{h}{\lambda_{1}}\right)^{2}+\left(\frac{h}{\lambda_{2}}\right)^{2}}$

$\frac{1}{\lambda^{2}}=\frac{1}{\lambda_{1}^{2}}+\frac{1}{\lambda_{2}^{2}}$

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