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Q.The de-Broglie wavelength of a particle of mass mm and kinetic energy KK is given by:

a
λ=h2mK\lambda = \frac{h}{\sqrt{2mK}}
b
λ=2mKh\lambda = \frac{\sqrt{2mK}}{h}
c
λ=h2mK\lambda = \frac{h}{2mK}
d
λ=h2mK\lambda = h\sqrt{2mK}

Correct Answer: Option A

The correct solution involves applying the fundamental concept to derive the final value step by step...

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