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PhysicsModern PhysicsDe Broglie & Matter WavesEasy2 minPYQ_2015

PhysicsEasystatement

De-Broglie wavelength of an electron accelerated by a voltage of $50 V$ is close to $(|e| = 1.6 \times 10^{- 19} C, m_{e} = 9.1 \times 10^{- 31} k g, h = 6.6 \times 10^{- 34} J s)$

Options:

Answer:

Solution:

De-Broglie wavelength $λ$ in given by

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

and for electron $λ = \frac{12.27 \overset{°}{A}}{\sqrt{V}}$

$= \frac{12.27 \overset{°}{A}}{\sqrt{50}}$

$= 1.7$ $\overset{°}{A}$

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

⏱ 2mℹ️ Source: PYQ_2015

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