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JEE Main 2021 (26 Aug, Shift 2) · Matter Waves: de Broglie Wavelength

Matter Waves: de Broglie Wavelength question from JEE Main 2021 (26 Aug, Shift 2)

Q89JEE Main 2021 · 26 AUG · SHIFT 2Easy

Matter Waves: de Broglie Wavelength

The de-Broglie wavelength of a particle having kinetic energy E is $\lambda$ . How much extra energy must be given to this particle so that the de-Broglie wavelength reduces to 75% of the initial value?

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Answer: (b)

$\lambda \propto \dfrac{1}{\sqrt{E}} \Rightarrow E' = E\left(\dfrac{\lambda}{\lambda'}\right)^2$ .
$\lambda' = 0.75\lambda = \dfrac{3}{4}\lambda \Rightarrow E' = \dfrac{16}{9}E$ .
Extra energy $= \dfrac{16}{9}E - E = \dfrac{7}{9}E$ .
So the answer is (b) $\dfrac{7}{9}E$ .

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