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Neglecting the variation of mass with energy, the de Broglie wavelength of an electron moving with energy E will be proportional to

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$\sqrt E \$

E^3/2

E^-1/2

answer is A.

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Detailed Solution

\lambda = \frac{h}{{\sqrt {2mE} }}\,\,;\,\lambda \alpha \frac{1}{{\sqrt E }}\,\,;\lambda \alpha {E^{ - 1/2}}\

$λ = \frac{h}{\sqrt{2 m E}}; λ \propto \frac{1}{\sqrt{E}}; λ \propto E^{- 1 / 2}$

λ = h / √(2mE) ; λ ∝ 1 / √E ; λ ∝ E^−1/2

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