If the deBroglie wavelength of an electron is equal to 1 0 − times the wavelength of a photon of frequency 6 × 1 0 14 Hz , then the speed of electron is equal to : (Speed of light = 3 × 1 0 8 m / s ) Planck's constant = 6.63 × 1 0 − 34 J . s Mass of electron = 9.1 × 1 0 − 31 kg )

de-Broglie wavelength, λ = mv h = 1 0 − 3 ( 6 × 1 0 14 3 × 1 0 8 ) [ ∵ λ = v c ] v = 9.1 × 1 0 − 31 × 3 × 1 0 5 6.63 × 1 0 − 34 × 6 × 1 0 14 v = 1.45 × 1 0 6 m / s

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PhysicsModern PhysicsDe Broglie & Matter WavesEasy2 minPYQ_2019

PhysicsEasystatement

If the deBroglie wavelength of an electron is equal to $1 0^{-}$ times the wavelength of a photon of frequency $6 \times 1 0^{14}$ $Hz$ , then the speed of electron is equal to : (Speed of light $= 3 \times 1 0^{8} m / s)$ Planck's constant $= 6.63 \times 1 0^{- 34} J . s$ Mass of electron $= 9.1 \times 1 0^{- 31} kg)$

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de-Broglie wavelength, $λ = \frac{h}{mv} = 1 0^{- 3} (\frac{3 \times 1 0 ^{8}}{6 \times 1 0 ^{14}}) [∵ λ = \frac{c}{v}]$ $v = \frac{6.63 \times 1 0 ^{- 34} \times 6 \times 1 0 ^{14}}{9.1 \times 1 0 ^{- 31} \times 3 \times 1 0 ^{5}} v = 1.45 \times 1 0^{6} m / s$

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

⏱ 2mℹ️ Source: PYQ_2019

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