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PhysicsModern PhysicsDe Broglie & Matter WavesMedium2 minPYQ_2020

PhysicsMediumsingle choice

If the kinetic energy of a particle is increased to 16 times its previous value, the percentage change in the de-Broglie wavelength of the particle is

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Kinetic energy $k = \frac{p ^{2}}{2 m}$ where $p$ is momentum $∴ \frac{k _{2}}{k _{1}} = \frac{P _{2}^{2}}{P _{1}^{2}} ∴ \frac{P _{2}}{P _{1}} = \frac{k _{2}}{k _{1}} = 16 = 4$ de Broglie wavelength $λ = \frac{h}{P}$ $∴ \frac{λ _{2}}{λ _{1}} = \frac{P _{1}}{P _{2}} = \frac{1}{4} ∴ λ_{2} = \frac{λ _{1}}{4} ∴ λ_{1} - λ_{2} = \frac{3}{4} λ_{1}$

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

⏱ 2mℹ️ Source: PYQ_2020

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