Question 200: Dual Nature of Matter and Radiation

Question

Two particles move at right angle to each other.
Their de-Broglie wavelengths are $$\lambda _1$$ and $$\lambda _2$$
respectively. The particles suffer perfectly
inelastic collision. The de-Broglie wavelength
$$\lambda _2$$ of the final particle, is given by :

Accepted Answer

Let the two particles be moving along x-direction
and y-direction.
So, the net momentum initially is $$\sqrt {{{{h^2}} \over {\lambda _1^2}} + {{{h^2}} \over {\lambda _2^2}}} $$
and final momentum will be $${h \over \lambda }$$.
Applying momentum conservation,
$${h \over \lambda } = \sqrt {{{{h^2}} \over {\lambda _1^2}} + {{{h^2}} \over {\lambda _2^2}}} $$
$$ \Rightarrow $$ $${1 \over {{\lambda ^2}}} = {1 \over {\lambda _1^2}} + {1 \over {\lambda _2^2}}$$

Answer

$$\lambda  = {{{\lambda _1} + {\lambda _2}} \over 2}$$

Answer

$${1 \over {{\lambda ^2}}} = {1 \over {\lambda _1^2}} + {1 \over {\lambda _2^2}}$$

Answer

$$\lambda  = \sqrt {{\lambda _1}{\lambda _2}} $$

Answer

$${2 \over \lambda } = {1 \over {{\lambda _1}}} + {1 \over {{\lambda _2}}}$$

Dual Nature of Matter and Radiation

Two particles move at right angle to each other. Their de-Broglie wavelengths are $\lambda _1 and \lambda _2 respectively. The particles suffer perfectly inelastic collision. The de-Broglie wavelength \lambda _2$ of the final particle, is given by :

Choose the correct answer: