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PhysicsModern PhysicsDe Broglie & Matter WavesEasy2 minPYQ_2020

PhysicsEasysingle choice

A proton and an $α$ -particle are accelerated through the same potential difference $V .$ The ratio of their de-Broglie wavelength is

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Answer:

Solution:

$λ = \frac{h}{\sqrt{2 m q V}}$

As $V$ is same.

$λ \propto \frac{1}{\sqrt{m q}}$

$\frac{λ_{p}}{λ_{α}} = \frac{\sqrt{m_{α} q_{α}}}{\sqrt{m_{p} q_{p}}} = \frac{\sqrt{{4 m}_{p} \times 2 q_{p}}}{\sqrt{m_{p} q_{p}}}$

$= 2 \sqrt{2}$

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

⏱ 2mℹ️ Source: PYQ_2020

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