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PhysicsModern PhysicsDe Broglie & Matter WavesMedium2 minPYQ_2019

PhysicsMediumsingle choice

The potential difference $V$ required for accelerating an electron to have the de Broglie wavelength of $1 \overset{˚}{A}$ is

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Solution:

$de Broglie wavelength, λ = \frac{h}{p} = \frac{h}{2 m e V} V = \frac{h ^{2}}{2 m e λ ^{2}} V = \frac{( 6.63 \times 1 0 ^{- 34} ) ^{2}}{2 \times 9.1 \times 1 0 ^{- 31} \times 1.6 \times 1 0 ^{- 19} \times ( 1 0 ^{- 10} ) ^{2}} = 150.95 V \approx 150 V .$

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

⏱ 2mℹ️ Source: PYQ_2019

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