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PhysicsModern PhysicsDe Broglie & Matter WavesEasy2 minPYQ_2021

PhysicsEasystatement

An $α$ particle and a proton are accelerated from rest by a potential difference of $200 V$ . After this, their de Broglie wavelengths are $λ_{α}$ and $λ_{p}$ respectively. The ratio $\frac{λ_{p}}{λ_{α}}$ is :

Options:

Answer:

Solution:

$λ = \frac{h}{p} = \frac{h}{\sqrt{2 m q V}}$

$\Rightarrow \frac{λ_{p}}{λ_{α}} = \sqrt{\frac{m_{α} q_{α}}{m_{p} q_{p}}} = \sqrt{\frac{4 m_{p} \times 2 e}{m_{p} \times e}} = \sqrt{8} = 2 \sqrt{2}$ $= 2 \sqrt{2}$

$\Rightarrow \frac{λ_{p}}{λ_{α}} = 2 \times 1.4 = 2.8$

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

⏱ 2mℹ️ Source: PYQ_2021

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