Dual Nature of Matter

JEE Main level, single correct, medium

49 of 200

Question 49 of 200

The de-Broglie wavelength of a particle having kinetic energy EE is λ.\lambda . How much extra energy must be given to this particle so that the de-Broglie wavelength reduces to 7575% of the initial value?

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Answer: (B) 7E9\dfrac{7E}{9}

Recall the formula of de-Broglie wavelength, λ=hp\lambda =\dfrac{h}{p}, here pp is the linear momentum of the particle, now use the relation between linear momentum and kinetic energy, p=2Em⇒E∝1λ2p=\sqrt{2Em}\Rightarrow E\propto \dfrac{1}{{\lambda }^{2}}⇒E=kλ2\Rightarrow E=\dfrac{k}{{\lambda }^{2}}
andE′=16k9λ2⇒ΔE=7E9\text{and}{E}^{'}=\dfrac{16k}{9{\lambda }^{2}} \Rightarrow \Delta E=\dfrac{7E}{9}.

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