(B) The de-Broglie wavelength $\lambda$ is given by the formula $\lambda = \frac{h}{mv}$.Given: Mass $(m) = 10^{-6} \ kg$,Velocity $(v) = 10 \ ms^{-1}

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(B) The de-Broglie wavelength $\lambda$ is given by the formula $\lambda = \frac{h}{mv}$.Given: Mass $(m) = 10^{-6} \ kg$,Velocity $(v) = 10 \ ms^{-1}$,and Planck's constant $(h) = 6.63 \times 10^{-34} \ J \cdot s$.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 $10^{-6} \ kg$ moving with a velocity of $10 \ ms^{-1}$ is:

AIIMS 2001 All AIIMS

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

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AIIMS 2001 Paper All AIIMS years →

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The basis of the quantum mechanical model of an atom is:

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

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If $\lambda _L, \lambda _M$ and $\lambda _N$ are the wavelengths of an electron in $L, M, N$ energy levels of $H^{-}$ atom respectively,what is their decreasing order?

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