(B) According to de-Broglie's equation,$\lambda = \frac{h}{mv}$.Given:$h = 6.6 \times 10^{-34} \ Js$$m = 9.0 \times 10^{-31} \ kg$$v = 2.2 \times 10^6

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(B) According to de-Broglie's equation,$\lambda = \frac{h}{mv}$.Given:$h = 6.6 \times 10^{-34} \ Js$$m = 9.0 \times 10^{-31} \ kg$$v = 2.2 \times 10^6 \ ms^{-1}$Substituting the values:$\lambda = \frac{6.6 \times 10^{-34}}{9.0 \times 10^{-31} \times 2.2 \times 10^6}$$\lambda = \frac{6.6}{19.8} \times 10^{-34+31-6} \ m$$\lambda = 0.333 \times 10^{-9} \ m$Since $1 \ nm = 10^{-9} \ m$,the wavelength is $0.33 \ nm$.

Class 11 Chemistry Structure of Atom De Broglie's principle

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The wavelength associated with the electron moving in the first orbit of hydrogen atom with velocity $2.2 \times 10^6 \ ms^{-1}$ (in $nm$) is $\left(m_e=9.0 \times 10^{-31} \ kg, h=6.6 \times 10^{-34} \ Js\right)$

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A
$0.66$
B
$0.33$
C
$0.22$
D
$0.44$

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Structure of Atom

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De Broglie's principle

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The wave nature of an electron was first given by

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Which of the following relations is correct, if the wavelength $(\lambda)$ is equal to the distance travelled by the electron in one second? $h$ is the Planck's constant and $m$ is the mass of electron.

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If the mass of particle $B$ is four times the mass of particle $A$ and the velocity of particle $A$ is eight times the velocity of particle $B$,then the ratio of the de Broglie wavelength of particle $A$ to particle $B$ will be .......

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The velocity $v$ of a de Broglie wave is given by
$\left[ \begin{array}{c} u = \text{Frequency} \\ m = \text{mass} \\ C = \text{Velocity of light} \end{array} \right]$

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$C_{60}$ emerging from a source at a speed $(v)$ has a de Broglie wavelength of $11.0 \ \mathring{A}$. The value of $v$ (in $m \ s^{-1}$) is closest to
[Planck's constant $h = 6.626 \times 10^{-34} \ J \ s$]

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The wavelength associated with the electron moving in the first orbit of hydrogen atom with velocity $2.2 \times 10^6 \ ms^{-1}$ (in $nm$) is $\left(m_e=9.0 \times 10^{-31} \ kg, h=6.6 \times 10^{-34} \ Js\right)$

Similar Questions

The wave nature of an electron was first given by

Which of the following relations is correct, if the wavelength $(\lambda)$ is equal to the distance travelled by the electron in one second? $h$ is the Planck's constant and $m$ is the mass of electron.

If the mass of particle $B$ is four times the mass of particle $A$ and the velocity of particle $A$ is eight times the velocity of particle $B$,then the ratio of the de Broglie wavelength of particle $A$ to particle $B$ will be .......

The velocity $v$ of a de Broglie wave is given by$\left[ \begin{array}{c} u = \text{Frequency} \\ m = \text{mass} \\ C = \text{Velocity of light} \end{array} \right]$

$C_{60}$ emerging from a source at a speed $(v)$ has a de Broglie wavelength of $11.0 \ \mathring{A}$. The value of $v$ (in $m \ s^{-1}$) is closest to[Planck's constant $h = 6.626 \times 10^{-34} \ J \ s$]

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The velocity $v$ of a de Broglie wave is given by
$\left[ \begin{array}{c} u = \text{Frequency} \\ m = \text{mass} \\ C = \text{Velocity of light} \end{array} \right]$

$C_{60}$ emerging from a source at a speed $(v)$ has a de Broglie wavelength of $11.0 \ \mathring{A}$. The value of $v$ (in $m \ s^{-1}$) is closest to
[Planck's constant $h = 6.626 \times 10^{-34} \ J \ s$]