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PhysicsModern PhysicsDe Broglie & Matter WavesMedium2 minQB

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The magnitude of the de Broglie wavelength $(λ)$ of electron (e), proton (p), neutron (n) and $α$ particle $(α)$ , all having the same energy of 1 MeV, in increasing order will follow the sequence

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Solution:

Same energy: $λ = \frac{h}{2 m E} \propto \frac{1}{m}$ . With $m_{α} > m_{n} > m_{p} > m_{e}$ , the increasing order is $λ_{α}, λ_{n}, λ_{p}, λ_{e}$ .

Subject:PhysicsTopic:Modern PhysicsSubtopic:De Broglie & Matter Waves

⏱ 2mℹ️ Source: QB

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