An electron and an alpha particle are accelerated by the same potential difference. Let ${\lambda }_{e}$ and ${\lambda }_{\alpha }$ denote the de Broglie wavelengths of the electron and the alpha particle, respectively, then:

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Question

An electron and an alpha particle are accelerated by the same potential difference. Let ${\lambda }_{e}$ and ${\lambda }_{\alpha }$ denote the de Broglie wavelengths of the electron and the alpha particle, respectively, then:

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YESPYQ · Accepted Answer

Correct answer: a) ${\lambda }_{e}>{\lambda }_{\alpha }$. Broglie wavelength λ = h / √(2mqV). For same V, λ ∝ 1/√(mq). Alpha particle (m α , q α = 2e) vs Electron (me, qe = e). Since m α >> m e and q α > q e , the product m α q α is much larger than m e q e . λ e will be larger than λ α . (NEW NCERT 12 th Page No. 284, 285, 286)

An electron and an alpha particle are accelerated by the same potential difference. Let \({\lambda }_{e}\) and \({\lambd…

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