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PhysicsModern PhysicsDe Broglie & Matter WavesMedium2 minQB

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A proton and an alpha particle are accelerated through the same potential difference. The ratio of their de Broglie wavelengths is:

Answer:

Solution:

$λ = \frac{h}{2 m q V}$ : same $V$ , so $\frac{λ _{p}}{λ _{α}} = \frac{m _{α} q _{α}}{m _{p} q _{p}} = 4 \times 2 = 22$ .

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

⏱ 2mℹ️ Source: QB

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