(B) The de-Broglie wavelength $(\lambda)$ is given by the formula: $\lambda = \frac{h}{mv}$ Where: $h$ (Planck's constant) = $6.63 \times 10^{-34} \ J

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(B) The de-Broglie wavelength $(\lambda)$ is given by the formula: $\lambda = \frac{h}{mv}$ Where: $h$ (Planck's constant) = $6.63 \times 10^{-34} \ J \cdot s$ $m$ (mass) = $10^{-6} \ kg$ $v$ (velocity) = $10 \ ms^{-1}$ Substituting the values: $\lambda = \frac{6.63 \times 10^{-34}}{10^{-6} \times 10} = \frac{6.63 \times 10^{-34}}{10^{-5}} = 6.63 \times 10^{-29} \ m$ Therefore,the correct option is $B$.

Class 11 Chemistry Structure of Atom De Broglie's principle

Medium

The de-Broglie wavelength associated with a particle of mass $10^{-6} \ kg$ with a velocity of $10 \ ms^{-1}$ is $:-$

A
$6.63 \times 10^{-22} \ m$
B
$6.63 \times 10^{-29} \ m$
C
$6.63 \times 10^{-31} \ m$
D
$6.63 \times 10^{-34} \ m$

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Structure of Atom

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De Broglie's principle

The diffraction of an electron beam indicates that:

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If the kinetic energy of an electron is $18.2 \times 10^{-25} \ J$,its de Broglie wavelength in $nm$ is: (mass of electron $= 9.1 \times 10^{-31} \ kg$; $h = 6.626 \times 10^{-34} \ J \ s$)

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When an electron in a hydrogen atom jumps from the third excited state to the ground state,the de-Broglie wavelength associated with the electron becomes

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Calculate the $\lambda$ of $CO_2$ molecule moving with a velocity $440 \ m/s.$

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The de-Broglie wavelength associated with a particle of mass $10^{-6} \ kg$ with a velocity of $10 \ ms^{-1}$ is $:-$

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