Question

Two particles of de Broglie wavelength $600nm$ and $300nm$ combine to form a particle (both particles were moving along same line and same direction). Then, the de Broglie wavelength of the new particle is

• A. 900 $nm$
• B. 180 $nm$
• C. 450 $nm$
• D. 200 $nm$

Answer 1

The correct option is D 200 $nm$

By conservation of momentum
$P=P_1+P_2\frac{h}{\lambda }=\frac{h}{{\lambda }_1}+\frac{h}{{\lambda }_2}$
$\frac{1}{\lambda }=\frac{1}{{\lambda }_1}+\frac{1}{{\lambda }_2}=\frac{1}{600}+\frac{1}{300}=\frac{3}{600}$
$\lambda =200nm$