Dual Nature of Matter

JEE Main level, single correct, medium

66 of 200

Question 66 of 200

A particle moving with kinetic energy EE has de Broglie wavelength λ\lambda . If energy ΔE\Delta E is added to its energy, the wavelength become λ2\dfrac{\lambda }{2} . Value of ΔE,\Delta E, is:

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Answer: (C) 3E3E

λ=h2(KE)m⇒λ∝1KE\lambda =\dfrac{h}{\sqrt{2\left(KE\right)m}}\Rightarrow \lambda \propto \dfrac{1}{\sqrt{KE}}
λλ2=KEfKEi\dfrac{\lambda }{\dfrac{\lambda }{2}}=\sqrt{\dfrac{KE{}_{f}}{K{E}_{i}}}
4KEi=KEf4K{E}_{i}=K{E}_{f}
⇒ΔE=4KEi−KEi=3KE=3E\Rightarrow \Delta E=4K{E}_{i}-K{E}_{i}=3KE=3E

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