Dual Nature of Matter

JEE Main level, single correct, medium

73 of 200

Question 73 of 200

Two particles move at right angle to each other. Their de Broglie wavelengths are λ1{\lambda }_{1} and λ2{\lambda }_{2} respectively. The particles suffer perfectly inelastic collision. The de Broglie wavelength λ\lambda of the final particle, is given by:

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Answer: (D) 1λ2=1λ12+1λ22\dfrac{1}{{\lambda }^{2}}=\dfrac{1}{{\lambda }_{1}^{2}}+\dfrac{1}{{\lambda }_{2}^{2}}

P1⃗=hλ1i^,P2⃗=hλ2j^\vec{{P}_{1}}=\dfrac{h}{{\lambda }_{1}}\hat{i}\text{,} \vec{{P}_{2}}=\dfrac{h}{{\lambda }_{2}}\hat{j}
P⃗=P1⃗+P2⃗=hλ1i^+hλ2j^\vec{P}=\vec{{P}_{1}}+\vec{{P}_{2}}=\dfrac{h}{{\lambda }_{1}}\hat{i}+\dfrac{h}{{\lambda }_{2}}\hat{j}
∣P⃗∣=h1λ12+1λ22\left|\vec{P}\right|=h\sqrt{\dfrac{1}{{\lambda }_{1}^{2}}+\dfrac{1}{{\lambda }_{2}^{2}}}
hλ=h1λ12+1λ22\dfrac{h}{\lambda }=h\sqrt{\dfrac{1}{{\lambda }_{1}^{2}}+\dfrac{1}{{\lambda }_{2}^{2}}}
1λ2=1λ12+1λ22\dfrac{1}{{\lambda }^{2}}=\dfrac{1}{{\lambda }_{1}^{2}}+\dfrac{1}{{\lambda }_{2}^{2}}

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