An electron is moving with a kinetic energy of 4.55 × 1 0 - 25 J. The de Broglie wavelength for it is (Given: h = 6.626 × 10 − 34 , m e = 9.1 × 10 − 31 kg )

K .E . = 1 2 mv 2 V = 2 K E m = 2 × 4.55 × 1 0 - 25 9.1 × 1 0 - 31 k g = 10 3 mS − 1 λ = h m v = 6.626 × 1 0 - 34 9.1 × 1 0 - 31 × 1 0 3 = 7.2 × 10 − 7 m

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Chemistry - Atomic Structure Question with Solution | TestHub

ChemistryAtomic StructureHeisensberg's Uncertainty PrincipleEasy2 minPYQ_2020

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An electron is moving with a kinetic energy of 4.55 $\times 1 0^{- 25}$ J. The de Broglie wavelength for it is
(Given: $h = 6.626 \times 10^{- 34}, m_{e} = 9.1 \times 10^{- 31} kg$ )

Options:

Answer:

Solution:

$K .E . = \frac{1}{2} {mv}^{2}$
$V = \sqrt{\frac{2 K E}{m}}$
$= \sqrt{\frac{2 \times 4.55 \times 1 0^{- 25}}{9.1 \times 1 0^{- 31} k g}}$
$= 10^{3} {mS}^{- 1}$
$λ = \frac{h}{m v} = \frac{6.626 \times 1 0^{- 34}}{9.1 \times 1 0^{- 31} \times 1 0^{3}}$
$= 7.2 \times 10^{- 7} m$

Stream:NTA_ABHYAS_NEETSubject:ChemistryTopic:Atomic StructureSubtopic:Heisensberg's Uncertainty Principle

⏱ 2mℹ️ Source: PYQ_2020

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