(D) According to the de Broglie equation,the wavelength $\lambda$ is given by $\lambda = \frac{h}{mv}$.Substituting the given values:$\lambda = \frac{

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(D) According to the de Broglie equation,the wavelength $\lambda$ is given by $\lambda = \frac{h}{mv}$.Substituting the given values:$\lambda = \frac{6.63 \times 10^{-34} \, J s}{(1.67 \times 10^{-27} \, kg) \times (1.0 \times 10^3 \, m s^{-1})}$$\lambda = \frac{6.63}{1.67} \times 10^{-34 + 27 - 3} \, m$$\lambda \approx 3.97 \times 10^{-10} \, m$To convert meters to nanometers $(1 \, nm = 10^{-9} \, m)$:$\lambda = 3.97 \times 10^{-10} \, m = 0.397 \times 10^{-9} \, m \approx 0.40 \, nm$.

Class 11 Chemistry Structure of Atom De Broglie's principle

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Calculate the wavelength in nanometers associated with a particle moving with a velocity of $1.0 \times 10^3 \, m s^{-1}$. (Given: mass $m = 1.67 \times 10^{-27} \, kg$ and Planck's constant $h = 6.63 \times 10^{-34} \, J s$)

A
$2.5$
B
$14$
C
$0.032$
D
$0.40$

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Structure of Atom

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De Broglie's principle

Calculate the de-Broglie wavelength of an electron residing in the $2$nd Bohr orbit of a hydrogen atom. (Bohr radius,$a_0 = 0.529 \ \mathring{A}$)

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The de Broglie wavelength of a tennis ball of $60 \ g$ moving with a velocity of $10 \ m/s$ is approximately

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Two particles $A$ and $B$ are in motion. If the wavelength associated with particle $A$ is $5 \times 10^{-8} \ m$,calculate the wavelength (in $\mathring{A}$) associated with particle $B$ if its momentum is half of $A$.

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If the de Broglie wavelength of an electron is $728.14 \ nm$,its kinetic energy in $J$ is: (mass of electron $= 9.1 \times 10^{-31} \ kg$; $h = 6.626 \times 10^{-34} \ J \ s$)

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The de Broglie wavelength is given as $1 \, \mathring{A}$ and the value of $h$ is $6.6252 \times 10^{-27} \, \text{erg} \cdot \text{s}$. What will be the momentum of the particle (in $\text{g} \cdot \text{cm/s}$)?

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Calculate the wavelength in nanometers associated with a particle moving with a velocity of $1.0 \times 10^3 \, m s^{-1}$. (Given: mass $m = 1.67 \times 10^{-27} \, kg$ and Planck's constant $h = 6.63 \times 10^{-34} \, J s$)

Similar Questions

Calculate the de-Broglie wavelength of an electron residing in the $2$nd Bohr orbit of a hydrogen atom. (Bohr radius,$a_0 = 0.529 \ \mathring{A}$)

The de Broglie wavelength of a tennis ball of $60 \ g$ moving with a velocity of $10 \ m/s$ is approximately

Two particles $A$ and $B$ are in motion. If the wavelength associated with particle $A$ is $5 \times 10^{-8} \ m$,calculate the wavelength (in $\mathring{A}$) associated with particle $B$ if its momentum is half of $A$.

If the de Broglie wavelength of an electron is $728.14 \ nm$,its kinetic energy in $J$ is: (mass of electron $= 9.1 \times 10^{-31} \ kg$; $h = 6.626 \times 10^{-34} \ J \ s$)

The de Broglie wavelength is given as $1 \, \mathring{A}$ and the value of $h$ is $6.6252 \times 10^{-27} \, \text{erg} \cdot \text{s}$. What will be the momentum of the particle (in $\text{g} \cdot \text{cm/s}$)?

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