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PhysicsModern PhysicsDe Broglie & Matter WavesEasy2 minPYQ_2019

PhysicsEasystatement

A particle $A$ of mass $m$ and charge $q$ is accelerated by a potential difference of $50 V .$ Another particle $B$ of mass $4 m$ and charge $q$ is accelerated by a potential difference of $2500 V .$ The ratio of de-Broglie wavelengths $\frac{λ_{A}}{λ_{B}}$ is close to:

Options:

Answer:

Solution:

de-Broglie wavelength $λ = \frac{h}{ρ}$

$= \frac{h}{\sqrt{2 m e V}}$

$\frac{λ_{A}}{λ_{B}} = \sqrt{\frac{4 m \times 2500}{m \times 50}}$

$= 2 \times 5 \sqrt{2}$

$= 14.14$

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

⏱ 2mℹ️ Source: PYQ_2019

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