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PhysicsModern PhysicsDe Broglie & Matter WavesEasy2 minPYQ_2022

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Electrons of mass $m$ with de-Broglie wavelength $λ$ fall on the target. The cut-off wavelength $λ_{0}$ is equal to [ $h =$ Planck's constant, $C =$ velocity of light $]$

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Solution:

Using de-Broglie equation $λ = \frac{h}{p}$ where $p = 2 m E$ $\Rightarrow λ = \frac{h}{2 m E}$ Energy of the X-ray emitted $E = \frac{h c}{λ _{0}}$ $∴ λ = \frac{h}{2 m \times \frac{h c}{λ _{0}}} \Rightarrow λ_{0} = \frac{2 m c λ ^{2}}{h}$

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

⏱ 2mℹ️ Source: PYQ_2022

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