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PhysicsModern PhysicsDe Broglie & Matter WavesMedium2 minPYQ_2024

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An electron and an alpha particle are accelerated by the same potential difference. Let $λ_{ε}$ and $λ_{a}$ denote the de-Broglie wavelengths of the electron and the alpha particle, respectively, then:NEET 2024 (Re-NEET)

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de-Broglie wavelength is given by $λ = \frac{h}{p} = \frac{h}{2 m q V}$ For same potential difference $λ \propto \frac{1}{m q}$ $\frac{λ _{α}}{λ _{e}} = \frac{m _{e} q _{e}}{m _{α} q _{α}}$ $∵ m_{α} ≫ m_{e}$ $λ_{e} > λ_{α}$

Stream:NEETSubject:PhysicsTopic:Modern PhysicsSubtopic:De Broglie & Matter Waves

⏱ 2mℹ️ Source: PYQ_2024

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